MCQ
$\operatorname{Lim}_{x \rightarrow 0} \frac{e-(1+2 x)^{\frac{1}{2 x}}}{x}$ is equal to :
- ✓$e$
- B$\frac{-2}{\mathrm{e}}$
- C$0$
- D$e-e^2$
$ =\operatorname{Lim}_{x \rightarrow 0}(-e) \frac{\left(e^{\frac{\ln (1+2 x)}{2 x}-1}-1\right)}{x} $
$ =\operatorname{Lim}_{x \rightarrow 0}(-e) \frac{\ln (1+2 x)-2 x}{2 x^2} $
$ =(-e) \times(-1) \frac{4}{2 \times 2}=e$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$A = \left[ {\begin{array}{*{20}{c}}
{{{10}^{30}} + 5}&{{{10}^{20}} + 4}&{{{10}^{20}} + 6}\\
{{{10}^4} + 2}&{{{10}^8} + 7}&{{{10}^{10}} + 2n}\\
{{{10}^4} + 8}&{{{10}^6} + 4}&{{{10}^{15}} + 9}
\end{array}} \right]$ ,
$n \in N$, then