Question

Let $f\left(x\right)=\frac{tanx}{x},thenlo{g}_e{lim}_{x\to 0}\left(\right[f\left(x\right)]+x^2{)}^{\frac{1}{\left\{f\right(x\left)\right\}}}$ is equal to, (where [.] denotes greatest integer function and { } fractional part)

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is C

3

${lim}_{x\to 0}\left[f\right(x\left)\right]={lim}_{x\to 0}\left[\frac{tanx}{x}\right]=1$
${lim}_{x\to 0}\left(\right[f\left(x\right)]+x^2{)}^{\frac{1}{\left(f\right(x\left)\right)}}={lim}_{x\to 0}(1+x^2{)}^{\frac{1}{\left(f\right(x\left)\right)}}\left(1^{\infty }form\right)$
Again, $f\left(x\right)=\frac{tanx}{x}=\frac{x+\frac{x^3}{3}+\frac{2}{15}+......}{x}$
$\left\{f\right(x\left)\right\}=\frac{x^2}{3}+\frac{2}{15}{x}^4+.......$
(i)becomes,
$lo{g}_e\left(e^{{lim}_{x\to 0}{x}^2\times \frac{1}{\left\{f\right(x\left)\right\}}}\right)=e{lim}_{x\to 0}\frac{x^2}{\frac{x^2}{3}+\frac{2}{15}{x}^4+...\infty }=3$