Continuity and Differentiability
Continuity via L'Hôpital — Finding f(0)
nta_pyq_2026_jan
Grade 12
Question:
If the function $f(x)=\dfrac{e^{x}\!\left(e^{\tan x-x}-1\right)+\log_e(\sec x+\tan x)-x}{\tan x-x}$ is continuous at $x=0$, then the value of $f(0)$ is equal to
$\dfrac{2}{3}$
$2$
$\dfrac{3}{2}$
$\dfrac{1}{2}$
Step-by-Step Solution
Key Concept: $f(0)=\lim_{x\to0}\dfrac{e^x(e^{\tan x-x}-1)+\ln(\sec x+\tan x)-x}{\tan x-x}$. Apply L'Hôpital repeatedly or split the limit.
$f(0)=3/2$.
Correct Answer: 3