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$$ \lim_{x \to \infty}(1+4/x)^\sqrt{x^2+1} $$ 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 ($e^4$).

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After replacing $\sqrt{x^2 + 1}$ by $x$, try raising to the power $\frac{1}{4} \cdot 4 = 1$ to get $$\lim_{x\to\infty} (1 + 4/x)^{x} = \lim_{x\to\infty} \left((1 + 4/x)^{x/4}\right)^{4} = \left(\lim_{x\to\infty} (1 + 4/x)^{x/4}\right)^4.$$ Then replace $x/4$ by $y$ to use your standard limit. –  TMM May 15 '12 at 13:50

3 Answers 3

HINT:

$x = \sqrt {x^2} < \sqrt{x^2 + 1} < \sqrt{x^2} + 1 =x + 1$

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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}}}} $.

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thank you for your reply –  Carmine Paternoster May 15 '12 at 12:11

$$\underset{\begin{smallmatrix} u\to 0 \\ v\to \infty \end{smallmatrix}}{\mathop{\lim }}\,{{\left( 1+u \right)}^{v}}={{\mathrm{e}}^{\underset{\begin{smallmatrix} u\to 0 \\ v\to \infty \end{smallmatrix}}{\mathop{\lim }}\,uv}}$$

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(-1) This is just wrong. Both sides are not well-defined, and this is a bad approach to "solving" the problem. –  TMM May 15 '12 at 13:52
    
TMM I think it is easy to prove $$ \underset{\begin{smallmatrix} u\to 0 \\ v\to \infty \end{smallmatrix}}{\mathop{\lim }}\,{{\left( 1+u \right)}^{v}}={{\mathrm{e}}^{\underset{\begin{smallmatrix} u\to 0 \\ v\to \infty \end{smallmatrix}}{\mathop{\lim }}\,uv}} $$so I don't give a proof.Here I present a wide range of formulas,it can solve the problem,and I think you can Understand it and prove it. –  Daoyi Peng May 16 '12 at 5:02

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