Evaluate $\lim\limits_{x \to \infty}\frac{(x+\sqrt{2})^{\sqrt{2}}-(x-\sqrt{2})^{\sqrt{2}}}{x^{\sqrt{2}-1}}$.

Question

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!

Answer 1

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

Answer 2

$$\lim\limits_{x \to \infty}\frac{(x+\sqrt{2})^{\sqrt{2}}-(x-\sqrt{2})^{\sqrt{2}}}{x^{\sqrt{2}-1}}=\lim\limits_{t \to 0}\frac{(1+\sqrt{2}t)^{\sqrt{2}}-(1-\sqrt{2}t)^{\sqrt{2}}}{t}\\ =2\sqrt2\lim\limits_{s \to 0}\frac{(1+s)^{\sqrt{2}}-(1-s)^{\sqrt{2}}}{2s}$$

is obviously the derivative of $2\sqrt2r^{\sqrt 2}$ taken at $r=1$, i.e. $\color{green}4$.

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It can also be justified by the generalized binomial theorem,

$$(1+s)^{\sqrt2}=1+\sqrt2s+\sqrt2(\sqrt2-1)\frac{s^2}2+\cdots$$

Answer 3

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

$$\lim_{x \to \infty}\left(\sqrt{2}x^{\sqrt{2} -1} \cdot \frac{2\sqrt{2}}{x^{\sqrt{2}-1}}\right) = \lim_{x\to\infty} 4 = 4$$