Limits MCQ Quiz in தமிழ் - Objective Question with Answer for Limits - இலவச PDF ஐப் பதிவிறக்கவும்

Concept:

Limit:

The number L is called the limit of function f(x) as x→a if and only if, for every ε > 0 there exists δ > 0 such that:

|f(x) - L| < ϵ 

Whenever 0 < |x - a| < δ 

Calculation:

Given that:

xn = 0.5xn−1 + 1 

$x_n=\frac{x_{n-1}}{2}+1$

x₁ = 1 (x₀ = 0 given)

$x_2=\frac{x_1}{2}+1=\frac{1}{2}+1=\frac{3}{2}$

$x_3=\frac{x_2}{2}+1=\frac{3}{2\times 2}+1=\frac{7}{4}$

$x_4=\frac{x_3}{2}+1=\frac{7}{4\times 2}+1=\frac{15}{8}$

$x_5=\frac{x_4}{2}+1=\frac{15}{8\times 2}+1=\frac{31}{16}$

The sequence is:

$=1,\frac{3}{2},\frac{7}{2^2},\frac{15}{2^3},\frac{31}{2^4},..,\frac{2^n-1}{2^{n-1}},..$

$\underset{n\to \mathrm{\infty }}{lim}{x}_n=\underset{n\to \mathrm{\infty }}{lim}\frac{2^n-1}{2^{n-1}}=\underset{n\to \mathrm{\infty }}{lim}\frac{2^n-1}{2^n}2$

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

Explanation:

$\mathrm{p}={\displaystyle \underset{\mathrm{x}\to \pi }{lim}\left(\frac{{\mathrm{x}}^2+\alpha \mathrm{x}+2{\pi }^2}{\mathrm{x}-\pi +2\sin \mathrm{x}}\right)}$

substitue x = π

$\mathrm{p}=\left(\frac{{\pi }^2+\alpha \pi +2{\pi }^2}{\pi -\pi +2\sin \pi }\right)$, sin π = 0

We can see that denominator becomes zero.

Thus for p to have finite value numerator is also zero.

π² + απ + 2π² = 0

α = - 3π

Now,  $\mathrm{p}=\left(\frac{{\pi }^2+\alpha \pi +2{\pi }^2}{\pi -\pi +2\sin \pi }\right)=\frac{0}{0}$form

Using L'Hospital's rule, we get

$\mathrm{p}={\displaystyle \underset{\mathrm{x}\to \pi }{lim}\left(\frac{2\mathrm{x}+\alpha }{1+2\cos \mathrm{x}}\right)}$

substituting x = π, α = - 3π

$\mathrm{p}=\left(\frac{2\pi -3\pi }{1+2\cos \pi }\right)$, cosπ = -1

p = $\frac{-\pi }{-1}$

p = π 

Concept:

L’ Hospital’s Rule:

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

This rule is only applicable for $\frac{0}{0}$ and $\frac{\mathrm{\infty }}{\mathrm{\infty }}$ indeterminate forms.

Calculation

As 

$\underset{x\to \mathrm{\infty }}{lim}\frac{xlnx}{1+x^2}=\frac{\mathrm{\infty }}{\mathrm{\infty }}$

Which is an indeterminate form.

Using L hospitality rule for $\underset{x\to \mathrm{\infty }}{lim}\frac{xlnx}{1+x^2}$

We get

$\underset{x\to \mathrm{\infty }}{lim}\frac{x\times \frac{1}{x}+lnx}{2x}=\underset{x\to \mathrm{\infty }}{lim}\frac{1+lnx}{2x}=\frac{\mathrm{\infty }}{\mathrm{\infty }}Indeterminateform$

Again applying the L hospitality rule, we get 

$\underset{x\to \mathrm{\infty }}{lim}\frac{\frac{1}{x}}{2}=\frac{\frac{1}{\mathrm{\infty }}}{2}=0$

∴ Value of 

$\underset{x\to \mathrm{\infty }}{lim}\frac{xlnx}{1+x^2}=0$

$lety=x^{\frac{1}{x}}$

Taking log_e on both sides

${\log }_{\mathrm{e}}\mathrm{y}=\frac{1}{\mathrm{x}}{\log }_{\mathrm{e}}\mathrm{x}$

$\mathrm{T}\mathrm{a}\mathrm{k}\mathrm{i}\mathrm{n}\mathrm{g}\underset{\mathrm{x}\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\mathrm{o}\mathrm{n}\mathrm{b}\mathrm{o}\mathrm{t}\mathrm{h}\mathrm{s}\mathrm{i}\mathrm{d}\mathrm{e}\mathrm{s}$

$\underset{x\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}{\log }_{\mathrm{e}}\mathrm{y}=\underset{x\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{1}{\mathrm{x}}{\log }_{\mathrm{e}}\mathrm{x}$

Substituting x = ∞

It is of the form $\frac{\mathrm{\infty }}{\mathrm{\infty }}$ 

Differentiating w.r.t x

$\underset{x\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}{\log }_{\mathrm{e}}\mathrm{y}=\underset{x\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{1}{\mathrm{x}}$

Substituting x = ∞

${\log }_{\mathrm{e}}\mathrm{y}=0\left(\frac{1}{\mathrm{\infty }}=0\right)$

∴ y = e⁰

$\underset{x\to 0}{lim}\frac{e^x-\left(1+x+\frac{x^2}{2}\right)}{x^3}=\frac{0}{0}$Not defined 

Using L – Hospital’s rule,

1^st derivation

$\underset{\mathrm{x}\to 0}{lim}\frac{\frac{d}{dx}\left[e^x-\left(1+x+\frac{x^2}{2}\right)\right]}{\frac{d}{dx}\left(x^3\right)}=\underset{\mathrm{x}\to 0}{lim}\frac{e^x-\left(0+1+\frac{2x}{2}\right)}{3x^2}=\frac{1-\left(0+1+0\right)}{0}=\frac{0}{0}$Not defined

2^nd derivation

$\underset{\mathrm{x}\to 0}{lim}\frac{{\mathrm{e}}^{\mathrm{x}}-\left(0+0+1\right)}{6\mathrm{x}}=\frac{0}{0}$Not defined

3^rd derivation

$\underset{\mathrm{x}\to 0}{lim}\frac{e^x-0}{6}=\frac{1}{6}$

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​​

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

Applying L’ Hospital  rule:

${\displaystyle \underset{x\to 0}{lim}\left({\displaystyle \frac{1-\cos x}{x^2}}\right)=\underset{x\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{\frac{d}{dx}(1-\cos x)}{\frac{d}{dx}(x^2)}=\underset{\mathrm{x}\to 0}{lim}\frac{sinx}{2x}}$

$\underset{\mathrm{x}\to 0}{lim}\frac{sinx}{2x}=\left(\frac{0}{0}\right)\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{m}$

Once again by L’ Hospital rule,

$\underset{\mathrm{x}\to 0}{lim}\frac{cos\mathrm{x}}{2}=\frac{1}{2}$

Concept:

$\begin{array}{c}lim\\ x\to 0\end{array}\frac{\sin x}{x}=1$

Calculation:

Given:

$\begin{array}{c}lim\\ x\to 0\end{array}\frac{\sin \left(\pi {\cos }^{2}x\right)}{x^2}$

cos² x = 1 – sin² x

$\begin{array}{c}lim\\ x\to 0\end{array}\frac{\sin \left(\pi {\cos }^{2}x\right)}{x^2}=\begin{array}{c}lim\\ x\to 0\end{array}\frac{\sin \left\{\pi \left(1-{\sin }^{2}x\right)\right\}}{x^2}$

$=\begin{array}{c}lim\\ x\to 0\end{array}\frac{\sin \left(\pi -\pi si{n}^{2}x\right)}{x^2}$

$=\begin{array}{c}lim\\ x\to 0\end{array}\frac{\sin \left(\pi si{n}^{2}x\right)}{x^2}$

Multiply and divide by π sin²x

$=\begin{array}{c}lim\\ x\to 0\end{array}\frac{\sin \left(\pi si{n}^{2}x\right)}{\pi {\sin }^{2}x}\times \frac{\pi {\sin }^{2}x}{x^2}$

$Sinceli{m}_{x\to 0}\frac{sinx}{x}=1$, the above expression gives:

Concept:

L’ Hospital’s Rule:

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

This rule is only applicable for $\frac{0}{0}$ and $\frac{\mathrm{\infty }}{\mathrm{\infty }}$ indeterminate forms.

Explanation:

$\underset{x\to 1}{lim}(\frac{1}{1nx}-\frac{1}{x-1})$

$\underset{x\to 1}{lim}\left[\frac{\left(x-1\right)-\ln x}{\ln x\left(x-1\right)}\right]$ which is in the form of 0/0

So, applying L- Hospital rule twice,

$\underset{x\to 1}{lim}\left[\frac{1-\frac{1}{x}}{\log x+\left(x-1\right)\left(\frac{1}{x}\right)}\right]$

then,

$\underset{x\to 1}{lim}\frac{\frac{1}{x^2}}{\frac{1}{x^2}+\frac{1}{x}}$

= $\frac{1}{1+1}=\frac{1}{2}=0.5$

$\underset{x\to \mathrm{\infty }}{lim}\left(ax+\left(\frac{7-\sqrt{3}{x}^2}{3-x}\right)\right)=b$

$ax+\frac{7-\sqrt{3}{x}^2}{3-x}$

$=\frac{ax\left(3-x\right)+7-\sqrt{3}{x}^2}{3-x}$

$=\frac{3ax-a{x}^2+7-\sqrt{3}{x}^2}{3-x}$

$=\frac{-\sqrt{3}{x}^2-a{x}^2+3ax+7}{3-x}$

$\underset{x\to \mathrm{\infty }}{lim}\frac{x^2\left(-\sqrt{3}-a\right)+x\left(3a+\frac{7}{x}\right)}{x\left(\frac{3}{x}-1\right)}$

∴ For the limit to be finite, the x² term in the numerator should be 0.
$\therefore -\sqrt{3}-a=0$

$a=-\sqrt{3}$

Now,

$\underset{x\to \mathrm{\infty }}{lim}\frac{x\left(-3\sqrt{3}+\frac{7}{x}\right)}{x\left(\frac{3}{x}-1\right)}$

$=\frac{-3\sqrt{3}}{-1}=3\sqrt{3}$

Concept:

A function f(x) is said to be continuous at a point x = a, in its domain if,

$\underset{\mathrm{x}\to \mathrm{a}}{lim}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{a}\right)$ exists or its graph is a single unbroken curve.

f(x) is Continuous at x = a

$\underset{\mathrm{x}\to {\mathrm{a}}^{+}}{lim}\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to {\mathrm{a}}^{-}}{lim}\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to \mathrm{a}}{lim}\mathrm{f}\left(\mathrm{x}\right)$

Calculation:

Given:

$f\left(x\right)=\frac{\left|x\right|}{x}$

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

$forx\to 0^{+},\left|x\right|=x$

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

$$Left hand limit and right hand limit are not equal 

∴ $\underset{x\to 0}{lim}f\left(x\right)$does not exist.

∴ All given options are true.