Given below are two statements : Statement I : lim x→0 (tan^(-1) x + loge √(1+x/1-x)-2x/x^5) = 2/5

Question

Given below are two statements :

Statement I : \(\lim\limits _{x \rightarrow 0}\left(\frac{\tan ^{-1} x+\log _{e} \sqrt{\frac{1+x}{1-x}}-2 x}{x^{5}}\right)=\frac{2}{5}\)

Statement II : \(\lim\limits _{\mathrm{x} \rightarrow 1}\left(\mathrm{x}^{\frac{2}{1-\mathrm{x}}}\right)=\frac{1}{\mathrm{e}^{2}}\)

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

(1) Statement I is false but Statement II is true

(2) Statement I is true but Statement II is false

(3) Both Statement I and Statement II are false

(4) Both Statement I and Statement II are true

Answer 1

Correct option is: (4) Both Statement I and Statement II are true 

\(\lim\limits _{x \rightarrow 0} \frac{\tan ^{-1} x+\frac{1}{2}[\ell n (1+x)-\ell n (1-x)]-2 x}{x^{5}}\)

\(=\lim\limits _{x \rightarrow 0} \frac{\left(x-\frac{x^{3}}{3}+\frac{x^{5}}{5} \ldots\right)+\frac{1}{2}\left[x-\frac{x^{2}}{2}+\frac{x^{3}}{3} \ldots-\left(-x-\frac{x^{2}}{2}-\frac{x^{3}}{3} \ldots\right)\right]-2 x}{x^{5}}\)

\(=\lim\limits _{x \rightarrow 0} \frac{2 x+\frac{2 x^{5}}{5} \ldots .-2 x}{x^{5}}=\frac{2}{5}\)

\(\lim\limits _{x \rightarrow 1} x^{\frac{2}{(1-x)}}=\mathrm{e}^{\lim\limits _{x \rightarrow 1}\left(\frac{2}{1-x}\right)(x-1)}=\mathrm{e}^{-2}\)

\(\Rightarrow\) Both statements correct