# Evaluate the following limits. \lim_{x\to\infty}(2+\frac{4}{x})^{2x}

Evaluate the following limits.
$\underset{x\to \mathrm{\infty }}{lim}{\left(2+\frac{4}{x}\right)}^{2x}$
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Given:
$\underset{x\to \mathrm{\infty }}{lim}{\left(2+\frac{4}{x}\right)}^{2x}$
Calculation:
$\underset{x\to \mathrm{\infty }}{lim}{\left(2+\frac{4}{x}\right)}^{2x}$ if we substitute limit $x\to \mathrm{\infty }$ in this we get indeterminate form ${2}^{\mathrm{\infty }}$
Let's define $y={\left(2+\frac{4}{x}\right)}^{2x}$
And take the log on both side and we will do some simplification
$\mathrm{ln}y={\mathrm{ln}\left(2+\frac{4}{x}\right)}^{2x}$
And take the log on both side and we will do some simplification
$\mathrm{ln}y={\mathrm{ln}\left(2+\frac{4}{x}\right)}^{2x}$
$⇒\mathrm{ln}y=2x\mathrm{ln}\left(2+\frac{4}{x}\right)$
$\mathrm{ln}y=\frac{\mathrm{ln}\left(2+\frac{4}{x}\right)}{\frac{1}{2x}}$
Taking limit $x\to \mathrm{\infty }$ on both side
$\underset{x\to \mathrm{\infty }}{lim}\left[\mathrm{ln}y\right]=\frac{\mathrm{ln}\left(2+\frac{4}{x}\right)}{\frac{1}{2x}}$
$\underset{x\to \mathrm{\infty }}{lim}\left[\mathrm{ln}y\right]=\frac{\mathrm{ln}\left(2+\frac{4}{\mathrm{\infty }}\right)}{\frac{1}{2\mathrm{\infty }}}$
$\underset{x\to \mathrm{\infty }}{lim}\left[\mathrm{ln}y\right]=\frac{\mathrm{ln}\left(2+0\right)}{0}=\frac{\mathrm{ln}\left(2\right)}{0}=\mathrm{\infty }$
We know that
$t={e}^{\mathrm{ln}y}$
This means we can do original limit as follow
$\underset{x\to \mathrm{\infty }}{lim}{\left(2+\frac{4}{x}\right)}^{2x}=\underset{x\to \mathrm{\infty }}{lim}\left[y\right]=\underset{x\to \mathrm{\infty }}{lim}{e}^{\mathrm{ln}y}={e}^{\underset{x\to \mathrm{\infty }}{lim}\left[\mathrm{ln}y\right]}{e}^{\mathrm{\infty }}=\mathrm{\infty }$
Therefore $\underset{x\to \mathrm{\infty }}{lim}{\left(2+\frac{4}{x}\right)}^{2x}=\mathrm{\infty }$