Question

$\underset{h\to 0}{\mathrm{Lim}}\frac{(e+h{)}^{\ln (e+h)}-e}{h}$ is

• A. 2
• B. 1
• C. 0
• D. non existent

Answer 1

Answer: (A)

Limit is $f^{\prime }(e)$ where $f(x)=x^{\ln x}=e^{{\ln }^{2}x}$
$\Rightarrow f^{\prime }(x)=e^{{\ln }^{2}x}\cdot \frac{2\ln x}{x}$
$f^{\prime }(e)=e\cdot \frac{2}{e}=2$