Given below are two statements: Statement I: _x1x^(2)/(1-x)=1e^2 Statement II: _x0(^-1x+12…

Question

Given below are two statements:

Statement I: $\displaystyle\lim_{x	o1}x^{\frac{2}{1-x}}=\dfrac{1}{e^2}$

Statement II: $\displaystyle\lim_{x	o0}\left(\frac{	an^{-1}x+\frac12\log_e\left(\frac{1+x}{1-x}ight)-2x}{x^5}ight)=\dfrac25$

In the light of the above statements, choose the correct answer from the options given below:

Accepted Answer

Statement I: Let $x=1+h$, $h\to0$: $x^{2/(1-x)}=(1+h)^{-2/h}\to e^{-2}=\dfrac1{e^2}$. True.

Statement II: Using $\tan^{-1}x=x-\dfrac{x^3}3+\dfrac{x^5}5-\cdots$ and $\dfrac12\log_e\left(\dfrac{1+x}{1-x}\right)=x+\dfrac{x^3}3+\dfrac{x^5}5+\cdots$:

$$\tan^{-1}x+\frac12\log_e\left(\frac{1+x}{1-x}\right)=2x+\frac{2x^5}{5}+\cdots$$

(the $x^3$ terms cancel). Subtracting $2x$ and dividing by $x^5$:
$$\lim_{x\to0}\frac{\frac{2x^5}{5}+\cdots}{x^5}=\frac25$$

True. Both statements are true.

Solutions · 1