Limits and Continuity Questions and Answers Pdf
Latest Limit and Continuity MCQ Objective Questions
Limit and Continuity MCQ Question 1:
What is \(\lim\limits_{x \to 0}\frac{sinx~+~log(1-x)}{x^2}\) equal to?
Answer (Detailed Solution Below)
Option 3 : -1/2
Concept:
L-Hospital Rule: Let f(x) and g(x) be two functions
Suppose that we have one of the following cases,
- \(\lim\limits_{x \to a} \frac{f(x)}{g(x)} =\frac{0}{0}\)
- \(\lim\limits_{x \to a} \frac{f(x)}{g(x)} =\frac{\infty }{\infty }\)
Then we can apply L-Hospital Rule ⇔\(\lim\limits_{x \to a} \frac{f(x)}{g(x)} =\lim\limits_{x \to a} \frac{f'(x)}{g'(x)}\)
Note: We have to differentiate both the numerator and denominator with respect to x unless and until
\(\lim\limits_{x \to a} \frac{f(x)}{g(x)} =l\neq \frac{0}{0}\) where l is a finite value.
Calculation:
\(\lim\limits_{x \to 0}\frac{sinx~+~log(1-x)}{x^2}\)
Form of limit\(\frac{0}{0}\)
Applying the L-Hospital rule,
\(\Rightarrow \displaystyle \lim_{ x\to 0}\frac{cosx+\frac{1}{1-x}(-1)}{2x}\)
\(\Rightarrow\displaystyle \lim_{ x\to 0}\frac{cosx-\frac{1}{1-x}}{2x}\)
\(\Rightarrow\displaystyle \lim_{ x\to 0}\frac{cosx-({1-x)^{-1}}}{2x}\)
Now Again Form of limit\(\frac{0}{0}\)
Applying the L-Hospital rule,
\(\Rightarrow\displaystyle \lim_{ x\to 0}\frac{-sinx-\frac{(1-x)-(-1)}{(1-x)^{2}}}{2}\)
\(\Rightarrow\displaystyle \lim_{ x\to 0}\frac{-sin~x-[(-1)(1-x)^{-2}(-1)]}{2}\)
\(\Rightarrow\displaystyle \lim_{ x\to 0}\frac{-sin~x-[(1-x)^{-2}]}{2}\)
\(\Rightarrow\displaystyle \lim_{ x\to 0}\frac{-sin~0-[(1-0)^{-2}]}{2}\)
=\(\frac{0-1}{2}\)
= -1/2
Limit and Continuity MCQ Question 2:
The value of k which makes \(f\left( x \right)\; = \;\left\{ {\begin{array}{*{20}{c}} {\frac{sin~5x}{x}\;,x \ne 0}\\ {k\;,x\; = \;0} \end{array}} \right.\) continuous at x = 0, is
Answer (Detailed Solution Below)
Option 4 : 5
Concept:
For Continuity of the function f(x), Left hand limit = Right hand limit
i.e.\(\displaystyle \lim_{ x\to 0^{-}}\)f(x) =\(\displaystyle \lim_{ x\to 0^{+}}\)f(x) =\(\displaystyle \lim_{ x\to 0^{}}f(x)\)
\(\displaystyle \lim_{ x\to 0^{}}\frac{sinx}{x}=1\).........(Equation 1)
Application:
For Continuity of the function f(x), Left hand limit = Right hand limit
i.e.\(\displaystyle \lim_{ x\to 0^{-}}\)f(x) =\(\displaystyle \lim_{ x\to 0^{+}}\)f(x) =\(\displaystyle \lim_{ x\to 0^{}}f(x)\)
\(\displaystyle \lim_{ h\to 0^{}}\frac{sin5 (0-h)}{0-h}=\displaystyle \lim_{ h\to 0^{}}\frac{sin5 (0+h)}{0-+h}\)=\(\displaystyle \lim_{ x\to 0^{}} f(x)\)
\(\displaystyle \lim_{ h\to 0^{}}\frac{sin5 (0-h)}{0-h} \times \frac{5}{5}=\displaystyle \lim_{ h\to 0^{}}\frac{sin5 (0+h)}{0+h}\times\frac{5}{5}\) = k
\(5\displaystyle \lim_{ h\to 0^{}}\frac{sin5h}{5h}\) =\(5\displaystyle \lim_{ h\to 0^{}}\frac{sin5h}{5h}\) = k (Using equation 1)
5(1) = k
\(\Rightarrow\)k=5
Limit and Continuity MCQ Question 3:
What is \(\lim\limits_{x \to 1}\frac{2(1-cosx)}{x^2}\) equal to?
Answer (Detailed Solution Below)
Option 2 : \(4sin^2\frac{1}{2}\)
Concept:
1-cosx=2\(sin^2\frac{x}{2}\)
Calculation:
In the given question, We have\(\displaystyle \lim_{ x \to 1} \frac{2(1-cosx)}{x^2}\)i.e.
\(\Rightarrow\) \(\displaystyle \lim_{ x \to 1} \frac{2(1-cosx)}{x^2}\)=\(\displaystyle \lim_{ x\to 1} \frac{(2)2sin^2\frac{x}{2}}{x^2}\)
\(\Rightarrow\) \(\displaystyle \lim_{ x \to 1} \frac{2(1-cosx)}{x^2} =\) \(\displaystyle \lim_{x \to 1} \frac {sin^2\frac{x}{2}}{\frac{x^2}{2^2}}\)
\(\Rightarrow\) \(\displaystyle \lim_{ x \to 1} \frac{2(1-cosx)}{x^2} =\) \(\frac{sin^2\frac{1}{2}}{\frac{1^2}{2^2}}\)
\(\Rightarrow\) \(\displaystyle \lim_{ x \to 1} \frac{2(1-cosx)}{x^2} =\) \(4sin^2\frac{1}{2}\)
Limit and Continuity MCQ Question 4:
The point(s), at which the function f given by f(x) = \(\left\{ \begin{array}{l} \frac{x}{{|x|}},x < 0\\ - 1,x \ge 0 \end{array} \right.\) is continuous, is/are :
Answer (Detailed Solution Below)
Option 1 : x ∈ R
Concept:
A function f(x) is continuous at x = a, if\(\lim_{x\to a^{-}}\) f(x) =\(\lim_{x\to a^{+}}\)f(x) = f(a).
Explanation:
Given, f(x) =\(\left\{ \begin{array}{l} \frac{x}{{|x|}},x < 0\\ - 1,x \ge 0 \end{array} \right.\)
We know that |x| = x when x > 0
and |x | = - x when x < 0
f(x) =\(\left\{ \begin{array}{l} \frac{x}{{(-x)}},x < 0\\ - 1,x \ge 0 \end{array} \right.\)
⇒ f(x) =\(\left\{ \begin{array}{l} -1,x < 0\\ - 1,x \ge 0 \end{array} \right.\)
Therefore, f(x) = -1
The function f(x) is a constant function, and a constant function is continuous for all x ∈ R.
The correct answer is option (1).
Limit and Continuity MCQ Question 5:
What is \(\displaystyle\lim_{h \rightarrow 0} \frac{\sin^2(x+h)−\sin^2x}{h}\) equal to ?
Answer (Detailed Solution Below)
Option 3 : sin 2x
Concept:
L-Hospital Rule: Let f(x) and g(x) be two functions
Suppose that we have one of the following cases,
I.\(\lim\limits_{x \to a} \frac{f(x)}{g(x)} =\frac{0}{0}\)
II.\(\lim\limits_{x \to a} \frac{f(x)}{g(x)} =\frac{\infty }{\infty }\)
Then we can apply L-Hospital Rule ⇔\(\lim\limits_{x \to a} \frac{f(x)}{g(x)} =\lim\limits_{x \to a} \frac{f'(x)}{g'(x)}\)
Note: We have to differentiate both the numerator and denominator with respect to x unless and until
\(\lim\limits_{x \to a} \frac{f(x)}{g(x)} =l\neq \frac{0}{0}\) where l is a finite value.
Solution:
\(\displaystyle\lim_{h \rightarrow 0} \frac{\sin^2(x+h)−\sin^2x}{h}\)
Form of limit\(\frac{0}{0}\)
Applying the L-Hospital rule,
=\(\displaystyle \lim_{h \to 0}\frac{2sin(x+h)cos(x+h)-0}{1}\)
= 2sin(x + 0)cos(x + 0)
= 2sinx.cosx
= sin2x
∴ The correct option is (3)
Top Limit and Continuity MCQ Objective Questions
Find the value of\(\rm \displaystyle \lim_{x \rightarrow \infty} 2x \sin \left(\frac{4} {x}\right)\)
Answer (Detailed Solution Below)
Option 3 : 8
Concept:
\(\rm \displaystyle \lim_{x \rightarrow 0} \frac{\sin x}{x} = 1\)
Calculation:
\(\rm \displaystyle \lim_{x \rightarrow \infty} 2x \sin \left(\frac{4} {x}\right)\)
=\(\rm 2 \times \displaystyle \lim_{x → ∞} \frac{\sin \left(\frac{4} {x}\right)}{\left(\frac{1}{x} \right )}\)
=\(\rm 2 \times \displaystyle \lim_{x → ∞} \frac{\sin \left(\frac{4} {x}\right)}{\left(\frac{4}{x} \right )} \times 4\)
Let\(\rm \frac {4}{x} = t\)
If x → ∞ then t → 0
=\(\rm 8 \times\displaystyle \lim_{t \rightarrow 0} \frac{\sin t}{t} \)
= 8 × 1
= 8
What is the value of\(\mathop {{\rm{lim}}}\limits_{x \to 0} \;\;\frac{{{{\left( {1 - \cos2x} \right)}^2}\;}}{{{x^4}}}\)
Answer (Detailed Solution Below)
Option 3 : 4
Concept:
- 1 - cos 2θ = 2 sin2 θ
- \(\mathop {{\rm{lim}}}\limits_{x \to 0} \;\;\frac{{\sin x}}{x} = 1\)
Calculation:
\(\mathop {{\rm{lim}}}\limits_{x \to 0} \;\;\frac{{{{\left( {1 - \cos2x} \right)}^2}\;}}{{{x^4}}}\)
=\(\;\mathop {{\rm{lim}}}\limits_{x \to 0} \;\;\frac{{{{\left( {2{{\sin }^2}x} \right)}^2}}}{{{x^4}}}\) (1 - cos 2θ = 2 sin2 θ)
=\(\;\mathop {{\rm{lim}}}\limits_{x \to 0} \;\;\frac{{4{{\sin }^4}x}}{{{x^4}}}\)
=\(\mathop {\lim }\limits_{x \to 0} 4\; × \;{\left( {\frac{{\sin x}}{x}} \right)^4}\)
= 4 × 1 = 4
Evaluate\(\rm \mathop {\lim }\limits_{x\rightarrow \infty} \frac{x}{\sqrt{1+2x^2}}\)
Answer (Detailed Solution Below)
Option 3 : \(\frac{1}{\sqrt 2}\)
Calculation:
We have to find the value of\(\rm \mathop {\lim }\limits_{x\rightarrow ∞} \frac{x}{\sqrt{1+2x^2}}\)
\(\rm \mathop {\lim }\limits_{x\rightarrow ∞} \frac{x}{\sqrt{1+2x^2}}\) [Form\(\frac{∞}{∞}\)]
This limit is of the form\(\frac{∞}{∞}\), Here, We can cancel a factor going to ∞ out of the numerator and denominator.
\(\rm \mathop {\lim }\limits_{x\rightarrow ∞} \frac{x}{\sqrt{1+2x^2}}\)
=\(\rm \mathop {\lim }\limits_{x\rightarrow \infty} \frac{x}{x\sqrt{\frac{1}{x^2}+2}}\)
Factor x becomes ∞ at x tends to ∞, So we need to cancel this factor from numerator and denominator.
=\(\rm \mathop {\lim }\limits_{x\rightarrow \infty} \frac{1}{\sqrt{\frac{1}{x^2}+2}}\)
=\(\frac{1}{\sqrt{\frac{1}{\infty^2}+2}}=\frac{1}{\sqrt{0+2}}=\frac{1}{\sqrt 2}\)
Evaluate\(\rm \mathop {\lim }\limits_{x\rightarrow 0} \frac{\log (1+2x)}{\tan 2x}\)
Answer (Detailed Solution Below)
Option 2 : 1
Concept:
\(\rm \mathop {\lim }\limits_{x\; \to \;a} \left[ {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right] = \frac{{\mathop {\lim }\limits_{x\; \to \;a} f\left( x \right)}}{{\mathop {\lim }\limits_{x\; \to \;a} g\left( x \right)}},\;provided\;\mathop {\lim }\limits_{x\; \to a} g\left( x \right) \ne 0\)
\(\rm \mathop {\lim }\limits_{x\; \to 0} {\frac{{\tan x}}{x}} = 1\)
\(\rm \mathop {\lim }\limits_{x\; \to 0} {\frac{{\log (1+x)}}{x}} = 1\)
Calculation:
\(\rm \mathop {\lim }\limits_{x\rightarrow 0} \frac{\log (1+2x)}{\tan 2x}\)
\(\rm = \mathop {\lim }\limits_{x\rightarrow 0} \frac{\frac{\log (1+2x)}{2x} \times 2x}{\frac{\tan 2x}{2x} \times 2x}\\= \frac{\mathop {\lim }\limits_{x\rightarrow 0}\frac{\log (1+2x)}{2x} }{\mathop {\lim }\limits_{x\rightarrow 0}\frac{\tan 2x}{2x} }\)
As we know\(\rm \mathop {\lim }\limits_{x\; \to 0} {\frac{{\tan x}}{x}} = 1\) and\(\rm \mathop {\lim }\limits_{x\; \to 0} {\frac{{\log (1+x)}}{x}} = 1\)
Therefore,\(\rm \mathop {\lim }\limits_{x\; \to 0} {\frac{{\tan 2x}}{2x}} = 1\) and\(\rm \mathop {\lim }\limits_{x\; \to 0} {\frac{{\log (1+2x)}}{2x}} = 1\)
Hence\(\rm \mathop {\lim }\limits_{x\rightarrow 0} \frac{\log (1+2x)}{\tan 2x} = \frac 1 1=1\)
Evaluate\(\rm \mathop {\lim }\limits_{x\rightarrow \infty} \frac{x^2}{{1+x^2}}\)
Answer (Detailed Solution Below)
Option 2 : 1
Calculation:
We have to find the value of\(\rm \mathop {\lim }\limits_{x\rightarrow \infty} \frac{x^2}{{1+x^2}}\)
\(\rm \mathop {\lim }\limits_{x\rightarrow \infty} \frac{x^2}{{1+x^2}}\) [Form\(\frac{∞}{∞}\)]
This limit is of the form\(\frac{∞}{∞}\), Here, We can cancel a factor going to ∞ out of the numerator and denominator.
\(\rm \mathop {\lim }\limits_{x\rightarrow \infty} \frac{x^2}{{1+x^2}}\)
=\(\rm \mathop {\lim }\limits_{x\rightarrow \infty} \frac{x^2}{x^2\left({\frac {1}{x^2}+1}\right)}\)
Factor x2 becomes ∞ at x tends to ∞, So we need to cancel this factor from numerator and denominator.
=\(\rm \mathop {\lim }\limits_{x\rightarrow \infty} \frac{1}{\left({\frac {1}{x^2}+1}\right)}\)
=\(\frac{1}{{\frac{1}{\infty^2}+1}}=\frac{1}{{0+1}}=1\)
Examine the continuity of a function f(x) = (x - 2) (x - 3)
Answer (Detailed Solution Below)
Option 3 : Continuous everywhere
Concept:
- We say f(x) is continuous at x = c if
LHL = RHL = value of f(c)
i.e.,\(\mathop {\lim }\limits_{{\rm{x}} \to {{\rm{c}}^ - }} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {{\rm{c}}^ + }} {\rm{f}}\left( {\rm{x}} \right) = {\rm{f}}\left( {\rm{c}} \right)\)
Calculation:
\(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} {\rm{f}}\left( {\rm{x}} \right)\; = \;\mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {\rm{a}}} \left( {{\rm{x}} - 2} \right)\left( {{\rm{x}} - 3} \right)\) (a ϵ Real numbers)
\(= \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {\rm{a}}} {\rm{\;}}{{\rm{x}}^2} - 3{\rm{x}} - 2{\rm{x}} + 6\)
\(= \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {\rm{a}}} {\rm{\;}}{{\rm{x}}^2} - 5{\rm{x}} + 6\)
\(= {{\rm{a}}^2} - 5{\rm{a}} + 6\)
∴ f(x) = f(a), So continuous at everywhere
Important tip:
Quadratic and polynomial functions are continuous at each point in their domain
\(\mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{\sin \left( {{{\rm{x}}^3}} \right)}}{{{{\rm{x}}^2}{\rm{\;sinx}}}}\) is equal to
Answer (Detailed Solution Below)
Option 3 : 1
Concept:
- \(\mathop {\lim }\limits_{{\rm{\alpha }} \to 0} \frac{{\sin \left( {\rm{\alpha }} \right)}}{{\rm{\alpha }}} = 1\) and \(\mathop {\lim }\limits_{{\rm{\alpha }} \to 0} \frac{{\rm{\alpha }}}{{\sin \left( {\rm{\alpha }} \right)}} = 1\)
- \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \left[ {{\rm{f}}\left( {\rm{x}} \right){\rm{\;}} \pm {\rm{\;g}}\left( {\rm{x}} \right)} \right] = {\rm{\;}}\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} {\rm{f}}\left( {\rm{x}} \right){\rm{\;}} \pm \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {\rm{a}}} {\rm{\;g}}\left( {\rm{x}} \right)\)
- \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \left[ {\frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}}} \right]{\rm{\;}} = \frac{{\mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {\rm{a}}} {\rm{\;f}}\left( {\rm{x}} \right)}}{{\mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {\rm{a}}} {\rm{\;g}}\left( {\rm{x}} \right)}}\)
Calculation:
To find: \(\mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{\sin \left( {{{\rm{x}}^3}} \right)}}{{{{\rm{x}}^2}{\rm{\;sinx}}}} = ?\)
Multiply and divide by x,
\(\mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{{\rm{x}}\sin \left( {{{\rm{x}}^3}} \right)}}{{{{\rm{x}}^3}{\rm{\;sinx}}}} = \mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{\sin \left( {{{\rm{x}}^3}} \right)}}{{{{\rm{x}}^3}}}\frac{{\rm{x}}}{{{\rm{sinx}}}}\)
\(= \mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{\sin \left( {{{\rm{x}}^3}} \right)}}{{{{\rm{x}}^3}}} \cdot \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to 0} \frac{{\rm{x}}}{{{\rm{sinx}}}}\)
\(= 1 \cdot 1 = 1\)
Hence, option (3) is correct.
If\(\rm f(x)=\left\{\begin{matrix}\rm \dfrac{\sin 3x}{e^{2x}-1}, &\rm x \ne 0\\\rm k-2, &\rm x=0 \end{matrix}\right.\) is continuous at x = 0, then k = ?
Answer (Detailed Solution Below)
Option 4 : \(\dfrac{7}{2}\)
Concept:
Definition:
- A function f(x) is said to be continuous at a point x = a in its domain, if\(\rm \displaystyle \lim_{x\to a}f(x)\) exists oror if its graph is a single unbroken curve at that point.
- f(x) is continuous at x = a ⇔\(\rm \displaystyle \lim_{x\to a^+}f(x)=\lim_{x\to a^-}f(x)=\lim_{x\to a}f(x)=f(a)\).
Formulae:
- \(\rm \displaystyle \lim_{x\to 0}\dfrac{\sin x}{x}=1\)
- \(\rm \displaystyle \lim_{x\to 0}\dfrac{e^x-1}{x}=1\)
Calculation:
Since f(x) is given to be continuous at x = 0,\(\rm \displaystyle \lim_{x\to 0}f(x)=f(0)\).
Also,\(\rm \displaystyle \lim_{x\to a^+}f(x)=\lim_{x\to a^-}f(x)\) because f(x) is same for x > 0 and x < 0.
\(\rm \therefore \displaystyle \lim_{x\to 0}f(x)=f(0)\)
\(\rm \Rightarrow \displaystyle \lim_{x\to 0}\dfrac{\sin 3x}{e^{2x}-1}=k-2\)
\(\rm \Rightarrow \displaystyle \lim_{x\to 0}\dfrac{\dfrac{\sin 3x}{3x}\times3x}{\dfrac{e^{2x}-1}{2x}\times2x}=k-2\)
\(\rm \Rightarrow \dfrac{3}{2}=k-2\)
\(\rm \Rightarrow k=\dfrac{7}{2}\).
\(\mathop {\lim }\limits_{n \to \infty } \frac{{{2^{n + 1}} + {3^{n + 1}}}}{{{2^n} + {3^n}}}\) equals
Answer (Detailed Solution Below)
Option 1 : 3
\(\mathop {\lim }\limits_{n \to \infty } \frac{{{2^{n + 1}} + {3^{n + 1}}}}{{{2^n} + {3^n}}} \)
This can be written as:
\(= \mathop {\lim }\limits_{n \to \infty } \frac{{{2^n}2 + {3^n}3}}{{{2^n} + {3^n}}}\)
Taking 3n common, we can write:
\( = \mathop {\lim }\limits_{n \to \infty } \frac{{{3^n}\left[ {2.{{\left( {\frac{2}{3}} \right)}^n} + 3} \right]}}{{{3^n}\left[ {{{\left( {\frac{2}{3}} \right)}^n} + 1} \right]}}\)
\( = \mathop {\lim }\limits_{n \to \infty } \frac{{2.{{\left( {\frac{2}{3}} \right)}^n} + 3}}{{\left[ {{{\left( {\frac{2}{3}} \right)}^n} + 1} \right]}}\)
Here\(\frac 2 3 < 1\)
So,\(\left[ \frac {2}{3}\right]^{\infty} = 0\)
\(= \frac{{0 + 3}}{{0 + 1}} = 3\)
What is\(\rm \displaystyle\lim_{x \rightarrow 0} \dfrac{3^x + 3^{-x}-2}{x}\) equal to ?
Answer (Detailed Solution Below)
Option 1 : 0
Concept:
\(\rm \mathop {\lim }\limits_{x\; \to \;a} \left[ {f\left( x \right) + g\left( x \right)} \right] = \;\mathop {\lim }\limits_{x\; \to \;a} f\left( x \right) + \;\mathop {\lim }\limits_{x\; \to \;a} g\left( x \right)\)
\(\rm \mathop {\lim }\limits_{x\; \to \;0} \dfrac { (a^x - 1) }{x} = \log a\)
log mn = n log m
Calculation:
\(\rm \displaystyle\lim_{x \rightarrow 0} \dfrac{3^x + 3^{-x}-2}{x}\\= \displaystyle\lim_{x \rightarrow 0} \dfrac{3^x -1+ 3^{-x}-1}{x}\\= \displaystyle\lim_{x \rightarrow 0} \dfrac{3^x -1}{x}+\displaystyle\lim_{x \rightarrow 0} \dfrac{3^{-x} -1}{x}\\= \displaystyle\lim_{x \rightarrow 0} \dfrac{3^x -1}{x}+\displaystyle\lim_{x \rightarrow 0} \dfrac{(3^{-1})^x -1}{x}\\= \log 3 + \log (3^{-1})\\= \log 3 - \log 3\\=0\)
If\(\rm f(x) = \left\{ \begin{matrix} \rm \frac{2x-\sin^{-1} x}{2x + \tan^{-1} x}; & \rm x \ne 0 \\ \rm K; & \rm x = 0 \end{matrix}\right.\) is a continuous function at x = 0, then the value of k is:
Answer (Detailed Solution Below)
Option 4 : None of these
Concept:
Definition:
- A function f(x) is said tobe continuous at a point x = a in its domain, if\(\rm \displaystyle \lim_{x\to a}f(x)\) existsoror if its graph is a single unbroken curve at that point.
- f(x) is continuous at x = a ⇔\(\rm \displaystyle \lim_{x\to a^+}f(x)=\lim_{x\to a^-}f(x)=\lim_{x\to a}f(x)=f(a)\).
Calculation:
For x ≠ 0, the given function can be re-written as:
\(\rm f(x) = \left\{ \begin{matrix} \rm \frac{2x-\sin^{-1} x}{2x + \tan^{-1} x}; & \rm x \ne 0 \\ \rm K; & \rm x = 0 \end{matrix}\right.\)
Since the equation of the function is same for x < 0 and x > 0, we have:
\(\rm \displaystyle \lim_{x\to 0^+}f(x)=\lim_{x\to 0^-}f(x)=\lim_{x\to 0}\frac{2x-\sin^{-1} x}{2x + \tan^{-1} x}\)
=\(\rm \displaystyle \lim_{x\to 0}\frac{2- \frac {\sin^{-1} x} x}{2 +\frac {\tan^{-1} x} x} = \frac {2-1}{2+1} = \frac 1 3\)
For the function to be continuous at x = 0, we must have:
\(\rm \displaystyle \lim_{x\to 0}f(x)=f(0)\)
⇒ K =\(\dfrac{1}{3}\).
The value of\(\rm \mathop {\lim }\limits_{x \to 0} \frac{{x}}{\cos x }\) is
Answer (Detailed Solution Below)
Option 2 : 0
Calculation:
Here we have to find the value of\(\rm \mathop {\lim }\limits_{x \to 0} \frac{{x}}{\cos x }\)
Let L =\(\rm \mathop {\lim }\limits_{x \to 0} \frac{{x}}{\cos x }\)
Now put the value of limit, we get
=\(\rm \frac {0}{\cos 0} = \frac 01=0\)
What is\(\rm \displaystyle\lim_{x\rightarrow 0} \dfrac{\sin x \log (1-x)}{x^2}\) equal to?
Answer (Detailed Solution Below)
Option 1 : -1
Concept:
\(\rm \mathop {\lim }\limits_{x\; \to \;a} \left[ {f\left( x \right) \cdot g\left( x \right)} \right] = \;\mathop {\lim }\limits_{x\; \to \;a} f\left( x \right) \cdot \;\mathop {\lim }\limits_{x\; \to \;a} g\left( x \right)\)
\(\rm \mathop {\lim }\limits_{x\; \to \;0} \dfrac {\sin x }{x} = 1\\\rm \mathop {\lim }\limits_{x\; \to \;0} \dfrac {\log (1+x) }{x}\)
Calculation:
We have to find the value of\(\rm \displaystyle\lim_{x\rightarrow 0} \dfrac{\sin x \log (1-x)}{x^2}\)
As we know,\(\rm \mathop {\lim }\limits_{x\; \to \;a} \left[ {f\left( x \right) \cdot g\left( x \right)} \right] = \;\mathop {\lim }\limits_{x\; \to \;a} f\left( x \right) \cdot \;\mathop {\lim }\limits_{x\; \to \;a} g\left( x \right)\)
\(\therefore \rm \displaystyle\lim_{x\rightarrow 0} \dfrac{\sin x \log (1-x)}{x^2}= \rm \mathop {\lim }\limits_{x\; \to \;0} \dfrac {\sin x }{x} × \mathop {\lim }\limits_{x\; \to \;0} \dfrac {\log (1-x) }{x}\)
= 1 × \(\rm \mathop {\lim }\limits_{x\; \to \;0} \dfrac {\log (1+(-x)) }{x}\)
= \(\rm \mathop {\lim }\limits_{x\; \to \;0} \dfrac {\log (1+(-x)) }{-(-x)}\)
=\(-1 × \rm \mathop {\lim }\limits_{x\; \to \;0} \dfrac {\log (1+(-x)) }{(-x)}\)
= -1 × 1
= -1
Let f(2) = 2 and f'(2) = 2. Then \(\mathop {\lim }\limits_{{\rm{x}} \to 2} \frac{{{\rm{xf}}\left( 2 \right) - 2{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{x}} - 2}}\) is given by
Answer (Detailed Solution Below)
Option 3 : -2
Concept:
- L' hospital's Rule: For (0/0), (∞ /∞) form of limits
If suppose,\(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = \frac{0}{0}{\rm{\;OR\;}}\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = \frac{{ \pm \infty }}{{ \pm \infty }}{\rm{\;\;}}\)
Then we apply L-Hospital's Rule,
\(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = \mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{f}}\left( {\rm{x}} \right)} \right)}}{{\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{g}}\left( {\rm{x}} \right)} \right)}} = \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f'}}\left( {\rm{x}} \right)}}{{{\rm{g'}}\left( {\rm{x}} \right)}}{\rm{\;}}\)
Differentiate the numerator and differentiate the denominator and then take limit
Calculation:
Given: f(2) = 2 and f'(2) = 2.
To find: \(\mathop {\lim }\limits_{{\rm{x}} \to 2} \frac{{{\rm{xf}}\left( 2 \right) - 2{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{x}} - 2}}{\rm{\;}}\)=?
Check the form by putting x = 2
\(\mathop {\lim }\limits_{{\rm{x}} \to 2} \frac{{{\rm{xf}}\left( 2 \right) - 2{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{x}} - 2}} = \frac{{2\left( 2 \right) - 2\left( 2 \right)}}{{2 - 2}} = \frac{0}{0}\)
Apply L' hospital's rule,
\(\mathop {\lim }\limits_{{\rm{x}} \to 2} \frac{{{\rm{xf}}\left( 2 \right) - 2{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{x}} - 2}} = \mathop {\lim }\limits_{{\rm{x}} \to 2} \frac{{\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{xf}}\left( 2 \right) - 2{\rm{f}}\left( {\rm{x}} \right)} \right)}}{{\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{x}} - 2} \right)}} = \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to 2} \frac{{{\rm{f}}\left( 2 \right) - 2{\rm{f'}}\left( {\rm{x}} \right)}}{1}\)
Differentiate the numerator and differentiate the denominator, we get
Now, take limit at x = 2
\(\mathop {{\rm{lim}}}\limits_{{\rm{x}} \to 2} = {\rm{f}}\left( 2 \right) - 2{\rm{f'}}\left( 2 \right)\)
\(= 2 - 2\left( 2 \right) = - 2\)
Option (3) is correct.
What is the value of\(\rm \mathop {{\rm{lim}}}\limits_{x \to 0} \;\;\frac{{{{\left( {1 - \cos2x} \right)}^3}\;}}{{{x^6}}}\)
Answer (Detailed Solution Below)
Option 3 : 8
Concept:
Formulas:
1 - cos 2θ = 2 sin2 θ
1 + cos 2θ = 2 cos2 θ
\(\rm \mathop {{\rm{lim}}}\limits_{x \to 0} \;\;\frac{{\sin x}}{x} = 1\)
Calculation:
To Find: Value of\(\rm \mathop {{\rm{lim}}}\limits_{x \to 0} \;\;\frac{{{{\left( {1 - \cos2x} \right)}^3}\;}}{{{x^6}}}\)
\(\rm \mathop {{\rm{lim}}}\limits_{x \to 0} \;\;\frac{{{{\left( {1 - \cos2x} \right)}^3}\;}}{{{x^6}}}\)
=\(\rm \mathop {{\rm{lim}}}\limits_{x \to 0} \;\;\frac{{{{\left( {2{{\sin }^2}x} \right)}^3}}}{{{x^6}}}\) (∵1 - cos 2θ = 2 sin2 θ)
=\(\rm \mathop {{\rm{lim}}}\limits_{x \to 0} \;\;\frac{{8{{\sin }^6}x}}{{{x^6}}}\)
=\(\rm \mathop {\lim }\limits_{x \to 0} 8\; \times \;{\left( {\frac{{\sin x}}{x}} \right)^6}\)
= 8 × 1
= 8
Evaluate\(\rm \displaystyle \lim_{x \rightarrow 0} \frac{1 - \cos 2x }{x^2} \)
Answer (Detailed Solution Below)
Option 3 : 2
Concept:
\(\rm \displaystyle \lim_{x \rightarrow 0} \frac{\sin x }{x} =1\)
Trigonometry formulas:
1 - cos 2x = 2sin2 x
1 + cos 2x = 2cos2 x
Calculation:
Here, we have to find the value of the limit\(\rm \displaystyle \lim_{x \rightarrow 0} \frac{1 - \cos 2x }{x^2} \)
As we know, 1 - cos 2x = 2sin2 x
\(\rm \displaystyle \lim_{x \rightarrow 0} \frac{1 - \cos 2x }{x^2} \)
=\(\rm \displaystyle \lim_{x \rightarrow 0} \frac{2\sin^2 x }{x^2} \)
= 2 ×\(\rm \displaystyle \lim_{x \rightarrow 0} \frac{\sin x }{x} × \lim_{x \rightarrow 0} \frac{\sin x }{x}\)
= 2 × 1 × 1
= 2
Function f(x) = tan(2x) will be discontinuous at x =
Answer (Detailed Solution Below)
Option 4 : (2n + 1) π/4
Concept:
- f(c) must be defined. You can't have a hole in the function (denominator is zero i.e. infinity)
- cos x = 0, where x = odd multiple of π/2 = (2n + 1) π/2
Calculation:
Here, \(\tan \left( {2{\rm{x}}} \right) = \frac{{\sin \left( {2{\rm{x}}} \right)}}{{\cos \left( {2{\rm{x}}} \right)}}\)
Check where denominator become zero
We know cos x = 0, where x = (2n + 1) π/2
So for cos 2x = 0 where, 2x = (2n + 1) π/2
x = (2n + 1) π/4
∴ Given function is discontinuous at x = (2n + 1) π/4
Hence, option (4) is correct.
Hint:
- When dealing with a rational expression in which both the numerator and denominator are continuous.
- The only points in which the rational expression will be discontinuous where denominator become zero.
\(\mathop {\lim }\limits_{{\rm{x}} \to 2} \sqrt {{{\rm{x}}^2} + 5} = ?\)
Answer (Detailed Solution Below)
Option 3 : √9
Concept:
- \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \sqrt[{\rm{n}}]{{{\rm{f}}\left( {\rm{x}} \right)}} = \sqrt[{\rm{n}}]{{\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} {\rm{f}}\left( {\rm{x}} \right)}}\)
- \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} {\rm{f}}\left( {\rm{x}} \right) + {\rm{g}}\left( {\rm{x}} \right) = \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {\rm{a}}} {\rm{f}}\left( {\rm{x}} \right) + \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {\rm{a}}} {\rm{g}}\left( {\rm{x}} \right)\)
Calculation:
To find:\(\mathop {\lim }\limits_{{\rm{x}} \to 2} \sqrt {{{\rm{x}}^2} + 5} \) = ?
\(\mathop {\lim }\limits_{{\rm{x}} \to 2} \sqrt {{{\rm{x}}^2} + 5} = \sqrt {\mathop {\lim }\limits_{{\rm{x}} \to 2}( {{\rm{x}}^2} + 5}) \)
\(= \sqrt {\mathop {\lim }\limits_{{\rm{x}} \to 2} {{\rm{x}}^2} + \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to 2} 5} = \sqrt {{2^2} + 5} \)
\(= \sqrt 9 \)
Hence, option (3) is correct.
Evaluate\(\mathop {\lim }\limits_{{\rm{x\;}} \to 0} \frac{{\rm \sqrt {a+x} - {\rm{\;}}\sqrt {\rm a - x} }}{\rm 4x}\)
Answer (Detailed Solution Below)
Option 4 : None of the above
Calculation:
We have to find the value of\(\mathop {\lim }\limits_{{\rm{x\;}} \to 0} \frac{{\rm \sqrt {a+x} - {\rm{\;}}\sqrt {\rm a - x} }}{\rm 4x}\)
Rationalise the numerator, we get
\(\mathop {\lim }\limits_{{\rm{x\;}} \to 0} \frac{{\rm \sqrt {a+x} - {\rm{\;}}\sqrt {\rm a - x} }}{\rm 4x} \times \frac{{\rm \sqrt {a+x} + {\rm{\;}}\sqrt {\rm a - x} }}{{\rm \sqrt {a+x} + {\rm{\;}}\sqrt {\rm a - x} }} \)
=\(\mathop {\lim }\limits_{{\rm{x\;}} \to 0} \frac{{\rm {a+x} - {\rm{\;}} {\rm (a - x )} }}{\rm 4x({\rm \sqrt {a+x} + {\rm{\;}}\sqrt {\rm a - x} } )} \)
=\(\mathop {\lim }\limits_{{\rm{x\;}} \to 0} \frac{\rm 2x}{\rm 4x({\rm \sqrt {a+x} + {\rm{\;}}\sqrt {\rm a - x} } )} \)
=\(\mathop {\lim }\limits_{{\rm{x\;}} \to 0} \frac{\rm 1}{\rm 2({\rm \sqrt {a+x} + {\rm{\;}}\sqrt {\rm a - x} } )} \)
=\(\frac{1}{{4 \rm \sqrt a }}\)
Find the value of\(\rm \displaystyle \lim_{n \rightarrow \infty} \frac{1-n^2} {\sum n}\)
Answer (Detailed Solution Below)
Option 1 : -2
Concept:
\(\rm \sum n = \frac{n(n+1)}{2}\)
Calculation:
\(\rm \displaystyle \lim_{n \rightarrow \infty} \frac{1-n^2} {\sum n}\)
=\(\rm \displaystyle \lim_{n \rightarrow \infty} \frac{1-n^2} { \left[\frac{n(n+1)}{2} \right ]}\)
=\(\rm \displaystyle \lim_{n \rightarrow \infty} \frac{2(1-n^2)} {n^2+n}\)
Divide both numerator and denominator by n2
=\(\rm \displaystyle \lim_{n \rightarrow \infty} \frac{2\left[\frac{1-n^2}{n^2} \right ]} {\left[\frac{n^2+n}{n^2} \right ]}\)
=\(\rm \displaystyle \lim_{n \rightarrow \infty} \frac{2\left[\frac{1}{n^2} - 1 \right ]} {1 + \frac{1}{n} }\)
=\(\rm \frac{2(0-1)}{(1+0)}\)
= -2
Source: https://testbook.com/objective-questions/mcq-on-limit-and-continuity--5eea6a1439140f30f369f1a7
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