$ \lim_{x \to \infty}(1+4/x)^\sqrt{x^2+1} $ is like
$ \lim_{x \to \infty}(1+1/x)^x = e $
I have replaced $\sqrt{x^2+1}$ by $x$ but I haven't got the expected result ($e^4$).
$ \lim_{x \to \infty}(1+4/x)^\sqrt{x^2+1} $ is like
$ \lim_{x \to \infty}(1+1/x)^x = e $
I have replaced $\sqrt{x^2+1}$ by $x$ but I haven't got the expected result ($e^4$).
HINT:
$x = \sqrt {x^2} < \sqrt{x^2 + 1} < \sqrt{x^2} + 1 =x + 1$
Note that $\sqrt{(x^{2}+1)}=|x|\sqrt{1+\frac{1}{x^{2}}}$ so $a^{\sqrt{(x^{2}+1)}}=(a^{|x|})^{\sqrt{1+\frac{1}{x^{2}}}} $.
$\underset{\begin{smallmatrix} u\to 0 \\ v\to \infty \end{smallmatrix}}{\mathop{\lim }}\,{{\left( 1+u \right)}^{v}}={{\mathrm{e}}^{\underset{\begin{smallmatrix} u\to 0 \\ v\to \infty \end{smallmatrix}}{\mathop{\lim }}\,uv}}$