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

Concept:

\(\displaystyle \lim _{n \rightarrow \infty} \frac{a_1+a_2+...+a_n}{n}\) = \(\displaystyle \lim _{n \rightarrow \infty} a_n\)

Explanation:

y = \(\displaystyle \lim _{n \rightarrow \infty} \frac{1}{n}\)\((1+\sqrt{2}+\sqrt[3]{3}+\ldots+\sqrt[n]{n})\) 

y = \(\displaystyle \lim _{n \rightarrow \infty} \frac{(1+\sqrt{2}+\sqrt[3]{3}+\ldots+\sqrt[n]{n})}{n}\) 

y = \(\displaystyle \lim _{n \rightarrow \infty} ​​\sqrt[n]{n}\) 

Taking log on both sides

log y = \(\displaystyle \lim _{n \rightarrow \infty} ​​\frac1n \log {n}\)

log y = \(\displaystyle \lim _{n \rightarrow \infty} {​​\frac1n \over 1}\) (Using L'hospital rule) 

log y = 0

y = e0 = 1

Hence option (2) is correct

Explanation:

\(\left( \frac{d^3y}{dx^3} \right)^2 + \left( \frac{dy}{dx} \right)^5 - 4 = 0 \)

1. Identify the order by determining the highest derivative present in the equation.  

2. Ensure the equation is a polynomial in derivatives to determine the degree.  

Hence Order= 3, Degree= 2.  

m +n=3+2 = 5

Hence 5 is the answer.

Given: 

We are given that for  \(n \geq 3 \) , a regular n -sided polygon  \(P_n \)  is circumscribed by a circle of radius \(R_n \) ,

and let \( r_n \)  be the radius of the circle inscribed in  \(P_n \) .

We need to find the value of:  \(\lim_{n \to \infty} \left( \frac{R_n}{r_n} \right)^{n^2} \)

Explanation:

1. For a regular n -sided polygon, as n increases,

the polygon approaches the shape of a circle.

In this limiting case,

both the circumscribed radius \( R_n \) and the inscribed radius  \(r_n \)  tend to the radius of the circle, say R ,

but they converge at different rates.

2. For a regular n -sided polygon:

The circumscribed radius  \(R_n \)  is approximately the same as the radius of the circle around the polygon.

The inscribed radius \( r_n \) is slightly smaller and tends to approach \(R_n \)  as n becomes large.

3. The ratio  \(\frac{R_n}{r_n} \)  approaches 1 as \(n \to \infty \) , but at a specific rate.

By approximation in trigonometric terms, we get that for large n :

\(\frac{R_n}{r_n} = 1 + \frac{\pi^2}{2n^2} \)
   

4. Now consider the expression \( \left( \frac{R_n}{r_n} \right)^{n^2} \) :

\(\left( \frac{R_n}{r_n} \right)^{n^2} = \left( 1 + \frac{\pi^2}{2n^2} \right)^{n^2} \)
   

5. Taking the limit as \( n \to \infty \) :

\(\lim_{n \to \infty} \left( 1 + \frac{\pi^2}{2n^2} \right)^{n^2} = e^{\frac{\pi^2}{2}} \)

Hence Option(2) is the correct answer.   

Concept:

(i) \(\lim_{n\to\infty} (u_n)^{\frac1n}\) = \(\lim_{n\to\infty} \frac{u_{n+1}}{u_n}\)

(ii) \(\lim_{n\to\infty} \frac1n\sum_{n=1}^{\infty}f(\frac rn)\) = \(\int_0^1f(x)dx\)

Explanation:

(1): let y = \(\displaystyle \lim _{n \rightarrow \infty}\)\(\left(\left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right) \ldots \ldots .\left(1+\frac{n}{n}\right)\right)^{\frac{1}{n}}\)

    So, log y = \(\displaystyle \lim _{n \rightarrow \infty}\)\(\frac1n\left(\log(1+\frac{1}{n})+\log(1+\frac{2}{n})+ \ldots\log(1+\frac{n}{n})\right)\)

                   = \(\displaystyle \lim _{n \rightarrow \infty}\)\(\frac1n\sum_{r=1}^{\infty}\log(1+\frac{r}{n})\)

                  = \(\int_0^1\log(1+x)dx\) (using concept (i))

                 = \([x.\log (1+x)]_0^1-\int_0^1\frac{1}{1+x}.xdx\) (by parts method)

                = log 2 - 0 - \(\int_0^1(1-\frac1{1+x})dx\)

                = log 2 - \([x-\log (1+x)]_0^1\)

               = log 2 - 1 + log 2 = 2log 2 - 1 = log 4 - log e = \(\log\frac4e\)

So log y = \(\log\frac4e\) ⇒ y = \(\frac{4}{e}\)

Hence (1) is correct

(2): \(\displaystyle \lim _{n \rightarrow \infty}\left(2 n_{C_n}\right)^{\frac{1}{n}}\) = \(\displaystyle \lim _{n \rightarrow \infty}\frac{^{2n+2}C_{n+1}}{^{2n}C_{n}}\) (using concet (i))

                          = \(\displaystyle \lim _{n \rightarrow \infty}\frac{\frac{(2n+2)!}{(n+1)!.(n+1)!}}{\frac{(2n)!}{n!.n!}}\)

                        = \(\displaystyle \lim _{n \rightarrow \infty}\frac{(2n+2)!}{(n+1)!.(n+1)!}.\frac{n!.n!}{(2n)!}\)

                      = \(\displaystyle \lim _{n \rightarrow \infty}\frac{(2n+2)(2n+1)}{(n+1)(n+1)}\)

                    = \(\displaystyle \lim _{n \rightarrow \infty}\frac{(2+\frac2n)(2+\frac 1n)}{(1+\frac1n)(1+\frac1n)}\) = 4

(2) is correct

(3): \(\displaystyle \lim _{n \rightarrow \infty}\left(\frac{3 n !}{(n !)^3}\right)^{\frac{1}{n}}\) = \(\displaystyle \lim _{n \rightarrow \infty}\frac{\frac{(3n+3)!}{((n+1)!)^3}}{\frac{(3n)!}{(n!)^3}}\)(using concept (ii))

                             = \(\displaystyle \lim _{n \rightarrow \infty}{\frac{(3n+3)!}{((n+1)!)^3}}.\frac{(n!)^3}{(3n)!}\)

                           = \(\displaystyle \lim _{n \rightarrow \infty}{\frac{(3n+3).(3n+2)(3n+1)}{(n+1)^3}}\) 

                          = \(\displaystyle \lim _{n \rightarrow \infty}{\frac{(3+\frac 3n).(3+\frac2n)(3+\frac1n)}{(1+\frac1n)^3}}\)= 27

(3) is correct

All options are corrct

Concept:

\(\displaystyle \lim _{n \rightarrow \infty} \frac{a_1+a_2+...+a_n}{n}\) = \(\displaystyle \lim _{n \rightarrow \infty} a_n\)

Explanation:

y = \(\displaystyle \lim _{n \rightarrow \infty} \frac{1}{n}\)\((1+\sqrt{2}+\sqrt[3]{3}+\ldots+\sqrt[n]{n})\) 

y = \(\displaystyle \lim _{n \rightarrow \infty} \frac{(1+\sqrt{2}+\sqrt[3]{3}+\ldots+\sqrt[n]{n})}{n}\) 

y = \(\displaystyle \lim _{n \rightarrow \infty} ​​\sqrt[n]{n}\) 

Taking log on both sides

log y = \(\displaystyle \lim _{n \rightarrow \infty} ​​\frac1n \log {n}\)

log y = \(\displaystyle \lim _{n \rightarrow \infty} {​​\frac1n \over 1}\) (Using L'hospital rule) 

log y = 0

y = e⁰ = 1

Hence option (2) is correct

Concept:

For limit evaluation of indeterminate forms, i.e. for \(\frac{0}{0}\) or \(\frac{\infty }{\infty }\) form, we apply L’ Hospital’s Rule as:

\(\mathop {\lim }\limits_{x \to a} \left\{ {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right\} = \mathop {\lim }\limits_{x \to a} \left\{ {\frac{{f'\left( x \right)}}{{g'\left( x \right)}}} \right\} \)

Calculation:

The given limit is,

\(\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{x} - \frac{1}{{sinx}}} \right) = \;\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin x ~- ~x}}{{xsin\;x}}} \right) \)

By putting a limit we get \(\frac{0}{0}\) form, therefore, by L'Hospital Rule:

\(\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{x} - \frac{1}{{sin~x}}} \right) = \;\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\cos x~ - ~1}}{{xcos\;x ~+ ~\sin x}}} \right) \)

Again putting the value of limit we get \(\frac{0}{0}\) form, therefore, again using L'Hospital Rule:

\(\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{x} - \frac{1}{{sinx}}} \right) = \;\mathop {\lim }\limits_{x \to 0} \left( {\frac{{ - \sin x}}{{x\left( { - \sin x} \right)~ +~ \cos x ~+~ \cos x}}} \right) \)

\( = \;\mathop {\lim }\limits_{x \to 0} \left( {\frac{{ - \sin x}}{{2\cos x -~~ xsin\;x}}} \right) \)

Now, by putting the limit, we get:

\(\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{x} - \frac{1}{{\sin x}}} \right) = \frac{0}{{2 - 0}} = 0 \)

\(\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{x} - \frac{1}{{sin~x}}} \right)~=~0 \)

Concept:

To solve \(\mathop {\lim }\limits_{x \to ∞ } \frac{{f\left( x \right)}}{{g\left( x \right)}}\) of this type, divide both numerator and denominator by the variable with the highest power.

Calculation:

Given:

\(\mathop {\lim }\limits_{x - ∞ } \frac{{{x^2}\;-\;5x\;+\;4}}{{4{x^2}\;+\;2x}}\)

Replacing x with ∞ in above expression, it is an indeterminate form \(\left( {\frac{∞ }{∞ }} \right).\)

Take out common the highest degree term from numerator and denominator.

\(\therefore \;\mathop {\lim }\limits_{x \to ∞ } \left( {\frac{{{x^2}\left[ {1\;-\;\frac{5}{x}\;+\;\frac{4}{{{x^2}}}} \right]}}{{{x^2}\left[ {4\;+\;\frac{2}{x}} \right]}}} \right)\)

\(\because\mathop {\lim }\limits_{x \to \infty } \frac{1}{x} = 0\)

\(\therefore\;\mathop {{\rm{lim}}}\limits_{x \to ∞ } \left( {\frac{{\left[ {1\;-\;0\;+\;0} \right]}}{{\left[ {4\;+\;0} \right]}}} \right)\)

\(\therefore\frac{1}{4}\)

Concept: 

L-Hospital Rule: Let f(x) and g(x) be two functions

Suppose that we have one of the following cases,

I.  \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = \frac{0}{0}\)

II. \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = \frac{\infty }{\infty }\)

Then we can apply L-Hospital Rule as:

\(\mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}\left( {\bf{x}} \right)}}{{{\bf{g}}\left( {\bf{x}} \right)}} = \mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}'\left( {\bf{x}} \right)}}{{{\bf{g}}'\left( {\bf{x}} \right)}}\)

Calculation:

Given:

\(\mathop {{\rm{lim}}}\limits_{x \to 0} \frac{{cos\;x\; -\;1}}{{sin\;x\; - \;x}} = \left( {\frac{0}{0}} \right)\) form

Applying L’ Hospital  rule:

\(\mathop {{\rm{lim}}}\limits_{x \to 0} \frac{{cos\;x - 1}}{{sin\;x-\;x}} =\mathop {{\rm{lim}}}\limits_{x \to 0} \frac{\frac{d}{dx}({cos\;x\; -\;1})}{{\frac{d}{dx}(sin\;x\;-\;x})} =\mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{-sin\;x {\rm{}}}}{{\cos {\rm{x}\;-\;1}}} \)

\(\mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{-sin\; {\rm{x}}}}{{\cos {\rm{x}\,-1\;}}} = \left( {\frac{0}{0}} \right){\rm{form}}\)

Once again by L’ Hospital rule,

\({\rm{}}\mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{-cos\; {\rm{x}}}}{{{\rm{-sin~x}}}} = \frac{-1}{0} = \infty{\rm{}}\)

Concept:

We know 

⇒ 1 - cos x = 2 sin²(x/2)​

⇒ \(\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1\)

Calculation:

\(\mathop {{\rm{lim}}}\limits_{x \to 0} \frac{{1 - {\rm{co}}{{\rm{s}}x}}}{{{x^2}}} = \mathop {{\rm{lim}}}\limits_{x \to 0} \frac{{2{{\sin }^2}\left( {\frac{x}{2}} \right)}}{{{x^2}}}\)

Multiply and divide the denominator by 4

\(\mathop {\lim }\limits_{x \to 0} \frac{{2 \times {{\sin }^2}\left( {\frac{x}{2}} \right)}}{{4 \times \left( {\frac{{{x^2}}}{4}} \right)}}\)= \(\frac{1}{2}\mathop {\lim }\limits_{x \to 0} {\left( {\frac{{\sin \left( {\frac{x}{2}} \right)}}{{\left( {\frac{x}{2}} \right)}}} \right)^2} = \frac{1}{2} \)

Concept:

Whenever the ∞ - ∞ form occurs then substitute x as (1/t) and then solve it.

Calculation:

\(\mathop {\lim }\limits_{x \to \infty } \sqrt {{x^2} + x - 1} - x\)

\(Let\;x = \frac{1}{t},\;So,\;as\;x \to \infty ,\;t \to 0\)

\(= \mathop {\lim }\limits_{t \to 0} \sqrt {\frac{1}{{{t^2}}} + \frac{1}{t} - 1} - \frac{1}{t}\)

\(= \mathop {\lim }\limits_{t \to 0} \frac{{\sqrt {1 + t - {t^2}} - 1}}{t}\)

Since the function has (0/0) form, we can apply L’ hospital rule,

\(= \mathop {\lim }\limits_{t \to 0} \frac{{\sqrt {1 + t - {t^2}} - 1}}{t}\)

\(= \mathop {\lim }\limits_{t \to 0} \left\{ {\frac{{\left[ {\frac{{\left( {1 - 2t} \right)}}{{2\sqrt {1 + t - {t^2}} }}} \right]}}{1}} \right\}\)

\(= \frac{1}{2}\)

Concept:

L-Hospital Rule: Let f(x) and g(x) be two functions

Suppose that we have one of the following cases,

I.  \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = \frac{0}{0}\)

II. \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = \frac{\infty }{\infty }\)

Then we can apply L-Hospital Rule

 \(\mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}\left( {\bf{x}} \right)}}{{{\bf{g}}\left( {\bf{x}} \right)}} = \mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}'\left( {\bf{x}} \right)}}{{{\bf{g}}'\left( {\bf{x}} \right)}}\)

Analysis:

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

\(y = \mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{x + 2 - 1}}{{x + 2}}} \right)^{2x + 1}}\) 

\(y = \mathop {\lim }\limits_{x \to \infty } {\left( {1 - \frac{1}{{\left( {x + 2} \right)}}} \right)^{2x + 1}}\) 

Taking log on both sides, we get

\(\log y = \mathop {\lim }\limits_{x \to \infty } \left( {2x + 1} \right)\log \left( {1 - \frac{1}{{x + 2}}} \right)\) 

\( = \mathop {\lim }\limits_{x \to \infty } \frac{{\log \left( {1 - \frac{1}{{x + 2}}} \right)}}{{\frac{1}{{\left( {2x + 1} \right)}}}}\) 

\(= \frac{0}{0}\) from

Using L Hospital’s rule we get

\(\log y = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{1}{{\left( {1 - \frac{1}{{x + 2}}} \right)}} \times \frac{1}{{{{\left( {x + 2} \right)}^2}}}}}{{ - \frac{2}{{{{\left( {2x + 1} \right)}^2}}}}}\) 

\( = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{{\left( {x + 2} \right)}}{{\left( {x + 1} \right)}} \times \frac{1}{{{{\left( {x + 2} \right)}^2}}}}}{{ - \frac{2}{{{{\left( {2x + 1} \right)}^2}}}}}\) 

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

\( = \mathop {\lim }\limits_{x \to \infty } - \frac{1}{2}\frac{{{{\left( {2 + \frac{1}{x}} \right)}^2}}}{{\left( {1 + \frac{1}{x}} \right)\left( {1 + \frac{2}{x}} \right)\;}}\) 

\(\log y = \frac{{ - 4}}{2}\)

y = e^-2

Concept:

\({1^2 + 2^2 + 3^2 + .....n^2} = \frac {n\ (n\ +\ 1)(2n\ +\ 1)}{6}\)

Calculation:

Given:

\(\rm \displaystyle\lim_{n \rightarrow \infty} \dfrac{1^2 + 2^2 + 3^2 + .....n^2}{n^3}\)

= \(\rm \displaystyle\lim_{n \rightarrow \infty} \dfrac{n\ (n\ +\ 1)(2n\ +\ 1)}{6\ n^3}\)

= \(\rm \displaystyle\lim_{n \rightarrow \infty} \dfrac{n\ (n\ +\ 1)(2n\ +\ 1)}{6\ n^3}\) 

= \(\rm \displaystyle\lim_{n \rightarrow \infty} \dfrac{(n\ +\ 1)(2n\ +\ 1)}{6\ n^2}\)

= \(\rm \displaystyle\lim_{n \rightarrow \infty} \dfrac{n^2(1\ +\ \frac {1}{n})(2\ +\ \frac {1}{n})}{6\ n^2}\) 

= \( {\left (1\ +\ \frac {1}{\infty}\right ) \left (2\ +\ \frac {1}{\infty }\right )}\ \frac 1 6\)

= \(\frac 1 3\)

Put x = 0 and then y = 0,

i) \(\mathop {\lim }\limits_{x \to 0 } \frac{{xy}}{{{x^2} + {y^2}}}\mathop {\lim }\limits_{y \to0 } \left( {\frac{0}{{{0^2} + {y^2}}}} \right) = 0\)

Put y = 0 and then x = 0)

ii) \(\mathop {\lim }\limits_{\begin{array}{*{20}{c}} {x \to 0 } \end{array}} \frac{{xy}}{{{x^2} + {y^2}}}\mathop {\lim }\limits_{x \to 0} \left( {\frac{0}{{{x^2} + 0}}} \right) = 0\)

Put y = mx,

iii) \(\mathop {\lim }\limits_{\begin{array}{*{20}{c}} {x \to 0 } \end{array}} \frac{{xy}}{{{x^2} + {y^2}}}\mathop {\lim }\limits_{x \to 0} \frac{{x\left( {mx} \right)}}{{{x^2} + {m^2}{x^2}}}\)

\(\mathop {\lim }\limits_{x \to 0 } \left( {\frac{m}{{1 + {m^2}}}} \right) = \frac{m}{{1 + {m^2}}}\)

It does not depend on limits, it depends on m. 

∴ Limits do not exist

Concept:

\(\mathop {\lim }\limits_{x \to 1} \left( {\frac{{1 - {e^{ - c\left( {1 - x} \right)}}}}{{1 - {xe^{ - c\left( {1 - x} \right)}}}}} \right)\) = 0/0 form

Hence apply L – Hospital rule

Calculation: 

Now,

By applying L – Hospital rule, i.e.

By differentiating both the numerator and denominator

\(\frac{{\frac{d}{{dx}}\left( {1 - {e^{ - c(1 - x)}}} \right)}}{{\frac{d}{{dx}}\left( {1 - x{e^{ - c(1 - x)}}} \right)}} = \frac{{\left( { - c{e^{ - c(1 - x)}}} \right)}}{{\left( { - cx{e^{ - c(1 - x)}} - {e^{ - c(1 - x)}}} \right)}} = \frac{{\left( {c{e^{ - c(1 - x)}}} \right)}}{{\left( {cx{e^{ - c(1 - x)}} + {e^{ - c(1 - x)}}} \right)}} = \frac{c}{{cx + 1}}\)

\(\therefore \mathop {\lim }\limits_{x \to 1} \frac{c}{{ 1 + cx}} = \frac{c}{{1 + c}}\)

Concept:

\(L{t_{x \to 0}}\frac{{\sin x}}{x} = 1\)

Calculation:

\(L{t_{x \to 0}}\left( {{x^2} - \frac{{\sin x}}{x}} \right) = 0 - 1 = - 1\;\;\;\;as\;L{t_{x \to 0}}\frac{{\sin x}}{x} = 1\)

Alternate solution

If limit x → 0 is put than it is a form of (0/0). Note that in case of (0/0) and (∞/∞) form, apply L-hospital rule.

\(\mathop {\lim }\limits_{x \to 0} \frac{{3{x^2} - \cos x}}{1} = 0 - 1 = - 1\)