Aspire Faculty ID #18036 · Topic: JEE Main 2026 (24 January Morning Shift) · Just now
JEE Main 2026 (24 January Morning Shift)

If the function $f(x)=\dfrac{e^x\left(e^{\tan x}-1\right)+\ln(\sec x+\tan x)-x}{\tan x-x}$ is continuous at $x=0$, then the value of $f(0)$ is equal to:

Solution

$f(0)=\lim_{x\to0}\dfrac{e^{\tan x}-e^x+\ln(\sec x+\tan x)-x}{\tan x-x}$

Applying L’Hospital rule

$f(0)=\lim_{x\to0}\dfrac{e^{\tan x}\sec^2x-e^x+\sec x-1}{\sec^2x-1}$

$=\lim_{x\to0}\dfrac{e^{\tan x}(\sec^2x-1)+(e^{\tan x}-e^x)+\sec x-1}{\tan^2x}$

$=\lim_{x\to0}\left(\dfrac{e^{\tan x}(e^{\tan x-x}-1)}{\tan^2x}+\dfrac{1}{\sec x+1}\right)$

$\Rightarrow f(0)=1+0+\frac{1}{2}=\frac{3}{2}$

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