Limits of the form (1 power infinity) - Practice Questions & MCQ

Limits of the form 1^∞ (1 power infinity)

To find the limit of the form $1^{\infty}$, we will use the following results
If $\lim _{x \rightarrow a} f(x)=\lim _{x \rightarrow a} g(x)=0$
Then,

$
\lim _{x \rightarrow a}[1+f(x)]^{\frac{1}{g(x)}}=e^{\lim _{x \rightarrow a} \frac{f(x)}{g(x)}}
$
Or
When, $\lim _{x \rightarrow a} f(x)=1$ and $\lim _{x \rightarrow a} g(x)=\infty$
Then,

$
\begin{aligned}
\lim _{x \rightarrow a}[f(x)]^{g(x)} & =\lim _{x \rightarrow a}[1+(f(x)-1)]^{g(x)} \\
& =e^{\lim _{x \rightarrow a}(f(x)-1) g(x)}
\end{aligned}
$

Proof: 

Let $L=\lim _{x \rightarrow a}[1+f(x)]^{\frac{1}{g(x)}}$
Taking logs of both sides

$
\begin{aligned}
& \log (L)=\log \left(\lim _{x \rightarrow a}[1+f(x)]^{\frac{1}{g(x)}}\right) \\
& \log (L)=\lim _{x \rightarrow a} \frac{1}{g(x)} \log (1+f(x)) \\
& \log (L)=\lim _{x \rightarrow a} \frac{1}{g(x)}\left(\frac{\log (1+f(x))}{f(x)}\right) \cdot f(x)
\end{aligned}
$
As $\mathrm{f}(\mathrm{x})$ is tending to 0 , so

$
\begin{aligned}
& \log (L)=\lim _{x \rightarrow a} \frac{f(x)}{g(x)} \\
& L=e^{\lim _{x \rightarrow a}} \frac{\frac{f(x)}{g(x)}}{2}
\end{aligned}
$

Some particular cases  

(a) $\lim _{x \rightarrow 0}(1+x)^{\frac{1}{x}}=e$
(b) $\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^x=e$
(c) $\lim _{x \rightarrow 0}(1+c x)^{\frac{1}{x}}=e^c$
(d) $\lim _{x \rightarrow \infty}\left(1+\frac{c}{x}\right)^x=e^c$