Evaluate $\lim\limits_{x \to \infty}\frac{(x+\sqrt{2})^{\sqrt{2}}-(x-\sqrt{2})^{\sqrt{2}}}{x^{\sqrt{2}-1}}$. Problem
Evaluate $$\lim\limits_{x \to \infty}\frac{(x+\sqrt{2})^{\sqrt{2}}-(x-\sqrt{2})^{\sqrt{2}}}{x^{\sqrt{2}-1}}.$$
Solution
Notice that $$\lim\limits_{x \to \infty}\frac{(x+\sqrt{2})^{\sqrt{2}}-(x-\sqrt{2})^{\sqrt{2}}}{x^{\sqrt{2}-1}}=\lim\limits_{x \to \infty}x \cdot \left[\left(1+\frac{\sqrt{2}}{x}\right)^{\sqrt{2}}-\left(1-\frac{\sqrt{2}}{x}\right)^{\sqrt{2}}\right].$$
According to Taylor's Formula $(1+x)^{\alpha}=1+\dfrac{\alpha}{1!}x+\mathcal{O}(x)$, we have $$\left(1+\frac{\sqrt{2}}{x}\right)^{\sqrt{2}}=1+\frac{2}{x}+\mathcal{O}\left(\frac{\sqrt{2}}{x}\right)$$and $$\left(1-\frac{\sqrt{2}}{x}\right)^{\sqrt{2}}=1-\frac{2}{x}+\mathcal{O}\left(\frac{\sqrt{2}}{x}\right).$$Therefore,$$\left(1+\frac{\sqrt{2}}{x}\right)^{\sqrt{2}}-\left(1-\frac{\sqrt{2}}{x}\right)^{\sqrt{2}}=\frac{4}{x}+\mathcal{O}\left(\frac{\sqrt{2}}{x}\right).$$As a result,
\begin{align*}
\lim\limits_{x \to \infty}\frac{(x+\sqrt{2})^{\sqrt{2}}-(x-\sqrt{2})^{\sqrt{2}}}{x^{\sqrt{2}-1}}&=\lim\limits_{x \to \infty}x \cdot \left[\left(1+\frac{\sqrt{2}}{x}\right)^{\sqrt{2}}-\left(1-\frac{\sqrt{2}}{x}\right)^{\sqrt{2}}\right]\\
&=\lim\limits_{x \to \infty}x \cdot \left[
\frac{4}{x}+\mathcal{O}\left(\frac{\sqrt{2}}{x}\right)\right]\\
&=4.
\end{align*}
Hope to see another solution. Thanks!
 A: Let $a=\sqrt2$. (Other values also work.)
Then we have
$$
\begin{aligned}
&\lim_{x\to\infty}
\frac{(x+a)^a-(x-a)^a}{x^{a-1}}
\\
&\qquad=
\lim_{x\to\infty}
a\cdot \frac xa\left(\ \left(1+\frac ax\right)^a-\left(1-\frac ax\right)^a\ \right)
\\
&\qquad\qquad\text{ after forced division with $x^a$ in both numerator and denominator,}
\\
&\qquad=
\qquad\lim_{y\to0}
a\cdot \frac 1y\left(\ \left(1+y\right)^a-\left(1-y\right)^a\ \right)
\\
&\qquad\qquad\text{ using the substitution $y=a/x$,}
\\
&\qquad=
\lim_{y\to0}
a\cdot \frac 1y\left(\ \left(1+ay+\dots\right)-\left(1-ay+\dots\right)\ \right)
\\
&\qquad\qquad\text{ using the Taylor expansion around zero and neglecting $O(y^2)$,}
\\
&\qquad=2a^2\ .
\end{aligned}
$$
Computer aid and confirmation:
sage: var('x');
sage: a = sqrt(2)
sage: limit( ( (x+a)^a-(x-a)^a ) / x^(a-1), x=oo )
4

A: $$\lim\limits_{x \to \infty}\frac{(x+\sqrt{2})^{\sqrt{2}}-(x-\sqrt{2})^{\sqrt{2}}}{x^{\sqrt{2}-1}}=\lim\limits_{t \to 0}\frac{(1+\sqrt{2}t)^{\sqrt{2}}-(1-\sqrt{2}t)^{\sqrt{2}}}{t}\\
=2\sqrt2\lim\limits_{s \to 0}\frac{(1+s)^{\sqrt{2}}-(1-s)^{\sqrt{2}}}{2s}$$
is obviously the derivative of $2\sqrt2r^{\sqrt 2}$ taken at $r=1$, i.e. $\color{green}4$.

It can also be justified by the generalized binomial theorem,
$$(1+s)^{\sqrt2}=1+\sqrt2s+\sqrt2(\sqrt2-1)\frac{s^2}2+\cdots$$
A: Let $f(x) = x^{\sqrt{2}}$. Let's accept for the moment(actually, Yves' answer justifies this statement) that
$$f'(x) \sim \frac{(x+\sqrt{2})^{\sqrt{2}}-(x-\sqrt{2})^{\sqrt{2}}}{2\sqrt{2}}$$
, reminiscent of the definition of the derivative.
Now, it then follows that 
$$\lim_{x \to \infty} \frac{(x+\sqrt{2})^{\sqrt{2}}-(x-\sqrt{2})^{\sqrt{2}}}{x^{\sqrt{2} - 1}} =  \lim_{x \to \infty} \left(\frac{(x+\sqrt{2})^{\sqrt{2}}-(x-\sqrt{2})^{\sqrt{2}}}{2\sqrt{2}} \cdot \frac{2\sqrt{2}}{x^{\sqrt{2}-1}}\right) = \lim_{x \to \infty} \left(f'(x)\cdot \frac{2\sqrt{2}}{x^{\sqrt{2}-1}}\right)$$
But $f'(x) = \sqrt{2}x^{\sqrt{2}-1}$, so the above simplifies to 
$$\lim_{x \to \infty}\left(\sqrt{2}x^{\sqrt{2} -1} \cdot \frac{2\sqrt{2}}{x^{\sqrt{2}-1}}\right) = \lim_{x\to\infty} 4 = 4$$
