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Continuity — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsContinuity2 Marks
MCQ
If the function $f(x)=\left\{\left[\begin{array}{cc}{\left[\tan \left(\frac{\pi}{4}+x\right)\right]^{1 / x}} & \text { for } x \neq 0 \\ K & \text { for } x=0\end{array}\right.\right.$ is continuous at $x=0$, then $K=$ ?
  • A
    $e$
  • B
    $e^{-1}$
  • ✓
    $e^2$
  • D
    $e^{-2}$
✓

Answer

Correct option: C.
$e^2$
(c) : We are given that $f(x)$ is continuous at $x=0$.
$\therefore f(0)=\lim _{x \rightarrow 0} f(x)$
$\Rightarrow K=\lim _{x \rightarrow 0}\left[\tan \left(\frac{\pi}{4}+x\right)\right]^{\frac{1}{x}}$
$=\lim _{x \rightarrow 0}\left(\frac{1+\tan x}{1-\tan x}\right)^{\frac{1}{x}}=\lim _{x \rightarrow 0} \frac{\left[(1+\tan x)^{\frac{1}{\tan x}}\right]^{\frac{\tan x}{x}}}{\left.(1-\tan x)^{-\frac{1}{\tan x}}\right]^{\frac{\tan x}{x}}}$ $=\frac{e}{e^{-1}}=e^2$

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