Evaluation of Limits MCQ Quiz - Objective Question with Answer for Evaluation of Limits - Download Free PDF

Concept:

If y = ln f(x) then \(\frac{dy}{dx}=\frac{f'(x)}{f(x)}\)
Calculation:

Given function is y = ln(e^mx + me^-mx)

Differentiating, we get

\(\frac{dy}{dx}=\frac{1}{e^{mx}+e^{-mx}}(me^{mx}-me^{-mx})\)

At x = 0, we have

\(\frac{dy}{dx}=\frac{1}{e^{0}+e^{0}}(me^{m0}-me^{-m0})\)

⇒ \(\frac{dy}{dx}=0\)

∴ For y = ln(emx + me-mx), \(\frac{dy}{dx}=0\) at x = 0.

Concept:

Chain Rule: 

• If y = f(g(x)), then dy/dx = f '(g(x) × g'(x)
• If y = p(x)/q(x), then dy/dx = \({(p'(x)q(x) - p(x) q'(x)) \over (q(x))^2}\)

Formula used:

• (log x)' = 1/x
• (xⁿ)' = n x^n-1

Calculation:

If  \(\rm y = \log \left( \frac{1 + \sqrt x}{1 - \sqrt x} \right)\),

then, \({dy\over dx} = {1\over \left( \frac{1 + \sqrt x}{1 - \sqrt x} \right)} \times {d\over dx}\left( \frac{1 + \sqrt x}{1 - \sqrt x} \right)\)

⇒ \({dy\over dx} = { \left( \frac{1 - \sqrt x}{1 + \sqrt x} \right)} \times \frac{({1\over 2\sqrt x})(1 - \sqrt x) - (1+\sqrt x)({-1\over 2\sqrt x})}{(1 - \sqrt x)^2} \)

⇒ \({dy\over dx} = { \left( \frac{1 - \sqrt x}{1 + \sqrt x} \right)} \times ({1\over 2\sqrt x})\frac{(1 - \sqrt x) + (1+\sqrt x)}{(1 - \sqrt x)^2} \)

⇒ \({dy\over dx} = { \left( \frac{1 - \sqrt x}{1 + \sqrt x} \right)} \times ({1\over 2\sqrt x})\frac{ 2}{(1 - \sqrt x)^2} \)

⇒ \({dy\over dx} = { \left( \frac{1 - \sqrt x}{1 + \sqrt x} \right)} \times ({1\over \sqrt x})\frac{ 1}{(1 - \sqrt x)^2} \)

⇒ \({dy\over dx} = { \left( \frac{1}{\sqrt x(1 + \sqrt x)(1 - \sqrt x)} \right)} \)

⇒ \({dy\over dx} = { \left( \frac{1}{\sqrt x (1 - x)} \right)} \)

∴ The correct answer is option (4).

Explanation:

Given:

L = \(\mathop {\lim }\limits_{x \to 2} {({x^3} - {x^2} - 8x + 13)^{\frac{1}{{{x^3} - 3{x^2} + 4}}}}\)

Taking log on both side we get

log L = \(\mathop {\lim }\limits_{x \to 2} {\frac{1}{{{x^3} - 3{x^2} + 4}}}\) log(x³ - x² - 8x + 13)  ----(1)

Taking RHS

Using L'hospital's Rule we get

= \(\mathop {\lim }\limits_{x \to 2} {\frac{1}{{{3x^2} - 6{x}}}}\times\frac{1}{{x^3} - {x^2} - 8x + 13}{({3x^2} - {2x} - 8)}\)

Again using L'hospital's Rule we get

= \(\mathop {\lim }\limits_{x \to 2} \frac{1}{(6x-6)({x^3} - {x^2} - 8x + 13)+({3x^2} - 6{x})({3x^2} - {2x} - 8)}{({6x} - {2})}\)

So we can see that L does not exist.

Calculation

\(\lim_{x\rightarrow \frac{\pi}{2}} \left(\frac{1}{x-\frac{\pi}{2}}\int^{\left(\frac{\pi}{2}\right)^3}_{x^3} cos\left(t^{\frac{1}{3}}\right)dt\right)\)

Using L’hopital rule 

\(\Rightarrow\lim _{x \rightarrow \frac{\pi^{-}}{2}} \frac{0-\cos x \times 3 x^2}{2\left(x-\frac{\pi}{2}\right)}\)

\(\Rightarrow\lim _{x \rightarrow \frac{\pi^{-}}{2}} \frac{\sin \left(x-\frac{\pi}{2}\right)}{2\left(x-\frac{\pi}{2}\right)} \times \frac{3 \pi^2}{4} \)

\(\Rightarrow\frac{3 \pi^2}{8} \)

Hence option 3 is correct

Formula used 

a³ - b³ = (a - b)(a² + ab + b²)

Calculation

Givn \(\lim_{x \to 12} \frac{x^3 - 1728}{x - 12}\)

x³ - 1728 = (x - 12)(x² + 12x + 144)

⇒  \(\lim_{x \to 12} \frac{(x - 12)(x^2 + 12x + 144)}{x - 12}\) 

⇒ \(\lim_{x \to 12} (x^2 + 12x + 144)\) 

Now, substituting x = 12 

⇒ 12² + 12 (12) + 144 = 144 + 144 + 144 = 432 

∴ The value of the limit is: 432

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

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 \(\rm \mathop {\lim }\limits_{x\rightarrow 0} \frac{\log (1+2x)}{\tan 2x} = \frac 1 1=1\)

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.

= \(\rm \mathop {\lim }\limits_{x\rightarrow \infty} \frac{1}{\sqrt{\frac{1}{x^2}+2}}\)

= \(\frac{1}{\sqrt{\frac{1}{\infty^2}+2}}=\frac{1}{\sqrt{0+2}}=\frac{1}{\sqrt 2}\)

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.

= \(\rm \mathop {\lim }\limits_{x\rightarrow \infty} \frac{1}{\left({\frac {1}{x^2}+1}\right)}\)

= \(\frac{1}{{\frac{1}{\infty^2}+1}}=\frac{1}{{0+1}}=1\)

Concept:​

• \(\rm \displaystyle \lim_{x\to0}\dfrac{\tan x}{x}=1\).
• \(\rm \dfrac{d}{dx}\tan x=\sec^2x\).
• \(\rm \dfrac{d}{dx}\sec x=\tan x\sec x\).
• \(\rm \dfrac{d}{dx}\left[f(x)\times g(x)\right]=f(x)\dfrac{d}{dx}g(x)+g(x)\dfrac{d}{dx}f(x)\).

Indeterminate Forms: Any expression whose value cannot be defined, like \(\dfrac00\), \(\pm\dfrac{\infty}{\infty}\), 0⁰, ∞⁰ etc.

• For the indeterminate form \(\dfrac 0 0\), first try to rationalize by multiplying with the conjugate, or simplify by cancelling some terms in the numerator and denominator. Else, use the L'Hospital's rule.
• L'Hospital's Rule: For the differentiable functions f(x) and g(x), the \(\rm \displaystyle \lim_{x\to c} \dfrac{f(x)}{g(x)}\), if f(x) and g(x) are both 0 or ±∞ (i.e. an Indeterminate Form) is equal to the \(\rm \displaystyle \lim_{x\to c} \dfrac{f'(x)}{g'(x)}\) if it exists.

Calculation:

\(\rm \displaystyle\lim_{x\to 0}\dfrac{\tan x - x}{x^2 \tan x}=\dfrac00\) is an indeterminate form. Let us simplify and use the L'Hospital's Rule.

\(\rm \displaystyle\lim_{x\to 0}\dfrac{\tan x - x}{x^2 \tan x}=\lim_{x\to 0}\left[\dfrac{\tan x - x}{x^3}\times\dfrac{x}{\tan x}\right]\).

We know that \(\rm \displaystyle\lim_{x\to 0}\dfrac{x}{\tan x}=1\), but \(\rm \displaystyle \lim_{x\to 0}\dfrac{\tan x - x}{x^3}\) is still an indeterminate form, so we use L'Hospital's Rule:

\(\rm \displaystyle \rm \displaystyle \lim_{x\to 0}\dfrac{\tan x - x}{x^3}=\lim_{x\to 0}\dfrac{\sec^2 x - 1}{3x^2}\), which is still an indeterminate form, so we use L'Hospital's Rule again:

\(\rm \displaystyle \lim_{x\to 0}\dfrac{\sec^2 x - 1}{3x^2}= \lim_{x\to 0}\dfrac{2\sec x(\sec x\tan x)}{6x}=\lim_{x\to 0}\dfrac{\sec^2 x\tan x}{3x}\), which is still an indeterminate form, so we use L'Hospital's Rule again:

\(\rm \displaystyle \lim_{x\to 0}\dfrac{\sec^2 x\tan x}{3x}=\lim_{x\to 0}\dfrac{\sec^2 x\sec^2 x+\tan x[2\sec x(\sec x \tan x)]}{3}=\dfrac{1}{3}\).

∴ \(\rm \displaystyle\lim_{x\to 0}\dfrac{\tan x - x}{x^2 \tan x}=1\times\dfrac{1}{3}=\dfrac{1}{3}\).

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:

\(\rm \displaystyle\lim_{x \rightarrow 0} \dfrac{3^x + 3^{-x}-2}{x}\\= \displaystyle\lim_{x \rightarrow 0} \dfrac{3^x -1+ 3^{-x}-1}{x}\\= \displaystyle\lim_{x \rightarrow 0} \dfrac{3^x -1}{x}+\displaystyle\lim_{x \rightarrow 0} \dfrac{3^{-x} -1}{x}\\= \displaystyle\lim_{x \rightarrow 0} \dfrac{3^x -1}{x}+\displaystyle\lim_{x \rightarrow 0} \dfrac{(3^{-1})^x -1}{x}\\= \log 3 + \log (3^{-1})\\= \log 3 - \log 3\\=0\)

Formula used:

\(\rm \displaystyle\lim_ {x\rightarrow 0}\left({\frac{sin\ x}{x}}\right)=1\)

Calculation:

\(\rm \displaystyle\lim_ {x\rightarrow 0}\left({\frac{\sqrt{1-cosx^2}}{(1-cosx)}}\right)\)

Since, 1 - cos 2θ = sin²θ

⇒ \(\rm \displaystyle\lim_ {x\rightarrow0}\left({\frac{√{2sin^2\frac{x^2}{2}}}{(2sin^2\frac{x}{2})}}\right)\)

⇒ \(\frac{1}{√ 2}\rm \displaystyle\lim_ {x\rightarrow 0}\left({\frac{{\frac{x^2}{2}\times sin\frac{x^2}{2}}}{(sin^2\frac{x}{2})\times\frac{x^2}{2}}}\right)\)

∴  \(\frac{2}{√ 2}\rm \displaystyle\lim_ {x\rightarrow 0}\left(\frac{sin\frac{x^2}{2}}{\frac{x^2}{2}}\right)\times \left(\frac{\frac{x}{2}}{sin\frac{x}{2}}\right)^2\) = √2

Concept:

\(\rm \displaystyle \lim_{x \rightarrow 0} \frac{\sin x}{x} = 1\)

Calculation:

\(\rm \displaystyle \lim_{x → ∞} x \sin \left(\frac{π} {x}\right)\)

= \(\rm \displaystyle \lim_{x → ∞} \frac{\sin \left(\frac{π} {x}\right)}{\left(\frac{1}{x} \right )}\)

= \(\rm \displaystyle \lim_{x → ∞} \frac{\sin \left(\frac{π} {x}\right)}{\left(\frac{π}{x} \right )} × π\)

Let \(\rm \frac {π}{x} = t\)

If x → ∞ then t → 0

= \(\rm \displaystyle \lim_{t \rightarrow 0} \frac{\sin t}{t} × π\)

= 1 × π 

= π 

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

\(\mathop {\lim }\limits_{n \to \infty } \frac{{{2^{n + 1}} + {3^{n + 1}}}}{{{2^n} + {3^n}}} \)

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, \(\left[ \frac {2}{3}\right]^{\infty} = 0\)

\(= \frac{{0 + 3}}{{0 + 1}} = 3\)