Question:

Let $\alpha$ and $\beta$ be the real numbers such that ,

${\lim\limits_{x\to 0}{\frac{1}{x^3}(\frac{\alpha}{2}\int_{0}^{x}\frac{1}{1-t^2}dt+\beta x \cos x)}=2}$

Answer: 2.40

Solution:

First we write the question correctly as below

${\lim\limits_{x\to 0} {\frac {{\frac{\alpha}{2}\int_{0}^{x}\frac{1}{1-t^2}dt+\beta x \cos x}}{x^3}=2}}$

Since the limit is in $\frac {0}{0}$ form therefore, Applying L’Hopital’s Rule, we get

${\lim\limits_{x\to 0} {\frac {{\frac{\alpha}{2} (\frac{1}{1-x^2})+\beta \cos x-\beta x \sin x}}{3 x^2}=2}}$

Now, put $x=0$ in the limit, if it is in the indeterminate form then only the limit will continue further, so make it in Indeterminate form.

Now on putting $x=0$ we get,

$\frac{\alpha}{2}+\beta =0$ therefore $\alpha=-2\beta$

…………….(1)

Now on putting value of $\alpha$ we get,

${\lim\limits_{x\to 0} {\frac {\beta {(x^2-1)^{-1}}+\beta \cos x-\beta x \sin x}{3 x^2}=2}}$

The limit is again in $\frac {0}{0}$ form therefore, again Applying L’Hopital’s Rule, we get

${\lim\limits_{x\to 0} {\frac {\beta {(-2x (x^2-1)^{-2}}- \sin x- \sin x-x \cos x)}{6 x}=2}}$

Question

Let $x_0$ be the real number such that $e^{x_0}+x_0=0$ . For a given real number $\alpha$ , define

$g(x)=\frac{3xe^x+3x-\alpha e^x-\alpha x}{3(e^x +1)}$

for all real numbers $x$ .

Then which one of the following statements is TRUE?

(A) For $\alpha = 2$ , $\lim\limits_{x \to x_0}=\left| \Large \frac{g(x)+e^{x_0}}{x-x_0} \right|=0$

(B) For $\alpha = 2$ , $\lim\limits_{x \to x_0}=\left| \Large \frac{g(x)+e^{x_0}}{x-x_0} \right|=1$

(C) For $\alpha = 3$ , $\lim\limits_{x \to x_0}=\left| \Large \frac{g(x)+e^{x_0}}{x-x_0} \right|=0$

(D) For $\alpha = 3$ , $\lim\limits_{x \to x_0}=\left| \Large \frac{g(x)+e^{x_0}}{x-x_0} \right|=\frac{2}{3}$