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

Concept:

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{a_1+a_2+...+a_n}{n}}$ = ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{a}_n}$

Explanation:

y = ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}}$$(1+\sqrt{2}+\sqrt[3]{3}+\dots +\sqrt[n]{n})$ 

y = ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{(1+\sqrt{2}+\sqrt[3]{3}+\dots +\sqrt[n]{n})}{n}}$ 

y = ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\sqrt[n]{n}}$ 

Taking log on both sides

log y = ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}\log n}$

log y = ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{\frac{1}{n}}{1}}$ (Using L'hospital rule) 

log y = 0

y = e0 = 1

Hence option (2) is correct

Explanation:

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

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  $n\ge 3$ , 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:  $\underset{n\to \mathrm{\infty }}{lim}{\left(\frac{R_n}{r_n}\right)}^{n^2}$

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 $n\to \mathrm{\infty }$ , 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 ${\left(\frac{R_n}{r_n}\right)}^{n^2}$ :

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

5. Taking the limit as $n\to \mathrm{\infty }$ :

$\underset{n\to \mathrm{\infty }}{lim}{(1+\frac{{\pi }^2}{2n^2})}^{n^2}=e^{\frac{{\pi }^2}{2}}$

Hence Option(2) is the correct answer.   

Concept:

(i) $\underset{n\to \mathrm{\infty }}{lim}(u_n{)}^{\frac{1}{n}}$ = $\underset{n\to \mathrm{\infty }}{lim}\frac{u_{n+1}}{u_n}$

(ii) $\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}\sum _{n=1}^{\mathrm{\infty }}f(\frac{r}{n})$ = ${\int }_0^{1}f(x)dx$

Explanation:

(1): let y = ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}}$${((1+\frac{1}{n})(1+\frac{2}{n})\dots \dots .(1+\frac{n}{n}))}^{\frac{1}{n}}$

    So, log y = ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}}$$\frac{1}{n}(\log (1+\frac{1}{n})+\log (1+\frac{2}{n})+\dots \log (1+\frac{n}{n}))$

                   = ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}}$$\frac{1}{n}\sum _{r=1}^{\mathrm{\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 - ${\int }_0^1(1-\frac{1}{1+x})dx$

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

               = log 2 - 1 + log 2 = 2log 2 - 1 = log 4 - log e = $\log \frac{4}{e}$

So log y = $\log \frac{4}{e}$ ⇒ y = $\frac{4}{e}$

Hence (1) is correct

(2): ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{\left(2n_{C_n}\right)}^{\frac{1}{n}}}$ = ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{{}^{2n+2}{C}_{n+1}}{{}^{2n}{C}_n}}$ (using concet (i))

                          = ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{\frac{(2n+2)!}{(n+1)!.(n+1)!}}{\frac{(2n)!}{n!.n!}}}$

                        = ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{(2n+2)!}{(n+1)!.(n+1)!}.\frac{n!.n!}{(2n)!}}$

                      = ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{(2n+2)(2n+1)}{(n+1)(n+1)}}$

                    = ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{(2+\frac{2}{n})(2+\frac{1}{n})}{(1+\frac{1}{n})(1+\frac{1}{n})}}$ = 4

(2) is correct

(3): ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{\left(\frac{3n!}{(n!{)}^3}\right)}^{\frac{1}{n}}}$ = ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{\frac{(3n+3)!}{((n+1)!{)}^3}}{\frac{(3n)!}{(n!{)}^3}}}$(using concept (ii))

                             = ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{(3n+3)!}{((n+1)!{)}^3}.\frac{(n!{)}^3}{(3n)!}}$

                           = ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{(3n+3).(3n+2)(3n+1)}{(n+1{)}^3}}$ 

                          = ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{(3+\frac{3}{n}).(3+\frac{2}{n})(3+\frac{1}{n})}{(1+\frac{1}{n}{)}^3}}$= 27

(3) is correct

All options are corrct

Concept:

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{a_1+a_2+...+a_n}{n}}$ = ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{a}_n}$

Explanation:

y = ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}}$$(1+\sqrt{2}+\sqrt[3]{3}+\dots +\sqrt[n]{n})$ 

y = ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{(1+\sqrt{2}+\sqrt[3]{3}+\dots +\sqrt[n]{n})}{n}}$ 

y = ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\sqrt[n]{n}}$ 

Taking log on both sides

log y = ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}\log n}$

log y = ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{\frac{1}{n}}{1}}$ (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 $\frac{\mathrm{\infty }}{\mathrm{\infty }}$ form, we apply L’ Hospital’s Rule as:

$\underset{x\to a}{lim}\left\{\frac{f\left(x\right)}{g\left(x\right)}\right\}=\underset{x\to a}{lim}\left\{\frac{f^{\prime }\left(x\right)}{g^{\prime }\left(x\right)}\right\}$

Calculation:

The given limit is,

$\underset{x\to 0}{lim}\left(\frac{1}{x}-\frac{1}{sinx}\right)=\underset{x\to 0}{lim}\left(\frac{\sin x-x}{xsinx}\right)$

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

$\underset{x\to 0}{lim}\left(\frac{1}{x}-\frac{1}{sinx}\right)=\underset{x\to 0}{lim}\left(\frac{\cos x-1}{xcosx+\sin x}\right)$

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

$\underset{x\to 0}{lim}\left(\frac{1}{x}-\frac{1}{sinx}\right)=\underset{x\to 0}{lim}\left(\frac{-\sin x}{x\left(-\sin x\right)+\cos x+\cos x}\right)$

$=\underset{x\to 0}{lim}\left(\frac{-\sin x}{2\cos x-xsinx}\right)$

Now, by putting the limit, we get:

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

$\underset{x\to 0}{lim}\left(\frac{1}{x}-\frac{1}{sinx}\right)=0$

Concept:

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

Calculation:

Given:

$\underset{x-\infty }{lim}\frac{x^2-5x+4}{4x^2+2x}$

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

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

$\therefore \underset{x\to \infty }{lim}\left(\frac{x^2\left[1-\frac{5}{x}+\frac{4}{x^2}\right]}{x^2\left[4+\frac{2}{x}\right]}\right)$

$\because \underset{x\to \mathrm{\infty }}{lim}\frac{1}{x}=0$

$\therefore \underset{x\to \infty }{\mathrm{l}\mathrm{i}\mathrm{m}}\left(\frac{\left[1-0+0\right]}{\left[4+0\right]}\right)$

$\therefore \frac{1}{4}$

Concept: 

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

Suppose that we have one of the following cases,

I.  $\underset{\mathrm{x}\to \mathrm{a}}{lim}\frac{\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{g}\left(\mathrm{x}\right)}=\frac{0}{0}$

II. $\underset{\mathrm{x}\to \mathrm{a}}{lim}\frac{\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{g}\left(\mathrm{x}\right)}=\frac{\mathrm{\infty }}{\mathrm{\infty }}$

Then we can apply L-Hospital Rule as:

$\underset{\mathbf{x}\to \mathbf{a}}{lim}\frac{\mathbf{f}\left(\mathbf{x}\right)}{\mathbf{g}\left(\mathbf{x}\right)}=\underset{\mathbf{x}\to \mathbf{a}}{lim}\frac{{\mathbf{f}}^{\prime }\left(\mathbf{x}\right)}{{\mathbf{g}}^{\prime }\left(\mathbf{x}\right)}$

Calculation:

Given:

$\underset{x\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{cosx-1}{sinx-x}=\left(\frac{0}{0}\right)$ form

Applying L’ Hospital  rule:

$\underset{x\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{cosx-1}{sinx-x}=\underset{x\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{\frac{d}{dx}(cosx-1)}{\frac{d}{dx}(sinx-x)}=\underset{\mathrm{x}\to 0}{lim}\frac{-sinx}{\cos \mathrm{x}-1}$

$\underset{\mathrm{x}\to 0}{lim}\frac{-sin\mathrm{x}}{\cos \mathrm{x}-1}=\left(\frac{0}{0}\right)\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{m}$

Once again by L’ Hospital rule,

$\underset{\mathrm{x}\to 0}{lim}\frac{-cos\mathrm{x}}{-\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}}=\frac{-1}{0}=\mathrm{\infty }$

Concept:

We know 

⇒ 1 - cos x = 2 sin²(x/2)​

⇒ $\underset{x\to 0}{lim}\frac{\sin x}{x}=1$

Calculation:

$\underset{x\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{1-\mathrm{c}\mathrm{o}\mathrm{s}x}{x^2}=\underset{x\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{2{\sin }^2\left(\frac{x}{2}\right)}{x^2}$

Multiply and divide the denominator by 4

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

Concept:

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

Calculation:

$\underset{x\to \mathrm{\infty }}{lim}\sqrt{x^2+x-1}-x$

$Letx=\frac{1}{t},So,asx\to \mathrm{\infty },t\to 0$

$=\underset{t\to 0}{lim}\sqrt{\frac{1}{t^2}+\frac{1}{t}-1}-\frac{1}{t}$

$=\underset{t\to 0}{lim}\frac{\sqrt{1+t-t^2}-1}{t}$

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

$=\underset{t\to 0}{lim}\frac{\sqrt{1+t-t^2}-1}{t}$

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

$=\frac{1}{2}$

Concept:

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

Suppose that we have one of the following cases,

I.  $\underset{\mathrm{x}\to \mathrm{a}}{lim}\frac{\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{g}\left(\mathrm{x}\right)}=\frac{0}{0}$

II. $\underset{\mathrm{x}\to \mathrm{a}}{lim}\frac{\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{g}\left(\mathrm{x}\right)}=\frac{\mathrm{\infty }}{\mathrm{\infty }}$

Then we can apply L-Hospital Rule

 $\underset{\mathbf{x}\to \mathbf{a}}{lim}\frac{\mathbf{f}\left(\mathbf{x}\right)}{\mathbf{g}\left(\mathbf{x}\right)}=\underset{\mathbf{x}\to \mathbf{a}}{lim}\frac{{\mathbf{f}}^{\prime }\left(\mathbf{x}\right)}{{\mathbf{g}}^{\prime }\left(\mathbf{x}\right)}$

Analysis:

$y=\underset{x\to \mathrm{\infty }}{lim}{\left(\frac{x+1}{x+2}\right)}^{2x+1}$

$y=\underset{x\to \mathrm{\infty }}{lim}{\left(\frac{x+2-1}{x+2}\right)}^{2x+1}$ 

$y=\underset{x\to \mathrm{\infty }}{lim}{\left(1-\frac{1}{\left(x+2\right)}\right)}^{2x+1}$ 

Taking log on both sides, we get

$\log y=\underset{x\to \mathrm{\infty }}{lim}\left(2x+1\right)\log \left(1-\frac{1}{x+2}\right)$ 

$=\underset{x\to \mathrm{\infty }}{lim}\frac{\log \left(1-\frac{1}{x+2}\right)}{\frac{1}{\left(2x+1\right)}}$ 

$=\frac{0}{0}$ from

Using L Hospital’s rule we get

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

$=\underset{x\to \mathrm{\infty }}{lim}\frac{\frac{\left(x+2\right)}{\left(x+1\right)}\times \frac{1}{{\left(x+2\right)}^2}}{-\frac{2}{{\left(2x+1\right)}^2}}$ 

$=\underset{x\to \mathrm{\infty }}{lim}\frac{{\left(2x+1\right)}^2}{-2\left(x+1\right)\left(x+2\right)}$ 

$=\underset{x\to \mathrm{\infty }}{lim}-\frac{1}{2}\frac{{\left(2+\frac{1}{x}\right)}^2}{\left(1+\frac{1}{x}\right)\left(1+\frac{2}{x}\right)}$ 

$\log y=\frac{-4}{2}$

y = e^-2

Concept:

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

Calculation:

Given:

${\displaystyle \underset{\mathrm{n}\to \mathrm{\infty }}{lim}{\displaystyle \frac{1^2+2^2+3^2+.....{\mathrm{n}}^2}{{\mathrm{n}}^3}}}$

= ${\displaystyle \underset{\mathrm{n}\to \mathrm{\infty }}{lim}{\displaystyle \frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{6{\mathrm{n}}^3}}}$

= ${\displaystyle \underset{\mathrm{n}\to \mathrm{\infty }}{lim}{\displaystyle \frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{6{\mathrm{n}}^3}}}$ 

= ${\displaystyle \underset{\mathrm{n}\to \mathrm{\infty }}{lim}{\displaystyle \frac{(\mathrm{n}+1)(2\mathrm{n}+1)}{6{\mathrm{n}}^2}}}$

= ${\displaystyle \underset{\mathrm{n}\to \mathrm{\infty }}{lim}{\displaystyle \frac{{\mathrm{n}}^2(1+\frac{1}{\mathrm{n}})(2+\frac{1}{\mathrm{n}})}{6{\mathrm{n}}^2}}}$ 

= $(1+\frac{1}{\mathrm{\infty }})(2+\frac{1}{\mathrm{\infty }})\frac{1}{6}$

= $\frac{1}{3}$

Put x = 0 and then y = 0,

i) $\underset{x\to 0}{lim}\frac{xy}{x^2+y^2}\underset{y\to 0}{lim}\left(\frac{0}{0^2+y^2}\right)=0$

Put y = 0 and then x = 0)

ii) $\underset{\begin{array}{c}x\to 0\end{array}}{lim}\frac{xy}{x^2+y^2}\underset{x\to 0}{lim}\left(\frac{0}{x^2+0}\right)=0$

Put y = mx,

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

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

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

∴ Limits do not exist

Concept:

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

Hence apply L – Hospital rule

Calculation: 

Now,

By applying L – Hospital rule, i.e.

By differentiating both the numerator and denominator

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

$\therefore \underset{x\to 1}{lim}\frac{c}{1+cx}=\frac{c}{1+c}$

Concept:

$L{t}_{x\to 0}\frac{\sin x}{x}=1$

Calculation:

$L{t}_{x\to 0}\left(x^2-\frac{\sin x}{x}\right)=0-1=-1asL{t}_{x\to 0}\frac{\sin x}{x}=1$

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.

$\underset{x\to 0}{lim}\frac{3x^2-\cos x}{1}=0-1=-1$