Find the $\lim_{x\to 1} x^{\frac{1}{x^2-1}}$ without applying L'Hôpital's rule

Question

Professor wants me to find the limit

$$ \lim_{x\to 1} x^{\frac{1}{x^2-1}} $$ without using L'Hôpital and even gave the following advice:

Rewrite the exponent as a product of sum and difference, make the following change of variables $h=\frac 1{x-1},$ when $x$ tends to $1,$ $h$ will tend to infinity. Add and subtract $1$ to the $x$ of the base and use the Euler number limit .

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Despite that, I was unable to find a coherent answer.

I fail to see how rewriting the exponent of $\lim_{x\to 1} x^{\frac{1}{x^2-1}}$ as $\frac{1}{(x-1)(x+1)}$ would help me solve this, even less so when I turn $\frac{1}{x-1} $ into $h.$

What am I missing here ?

Answer 1

I'd like to suggest you the fast-running substitution $u:=x^2-1\thinspace.$

Let $\thinspace x^2-1=u$ as $u\to 0$, then you have :

$$ \begin{align}\lim_{x\to 1}x^{\frac 1{x^2-1}}&=\lim_{x\to 1} {\left(x^2\right)}^{\frac{1}{2(x^2-1)}}\\ &=\lim_{u\to 0}\left(\left(1+u\right)^{\frac 1u}\right)^{\frac 12}\\ &=e^{\frac 12}=\sqrt e\thinspace .\end{align}$$

Because, $\thinspace \lim_{u\to 0}(1+u)^{\frac 1u}=e\thinspace $ by definition of the number $e\thinspace .$

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Also, using your professor's hint $u:=\frac {1}{x-1}$ as $|u|\to\infty$, you have :

$$ \begin{align}\lim_{x\to 1}x^{\frac 1{x^2-1}}&=\lim_{x\to 1}x^{\frac {1}{(x-1)(x+1)}}\\ &=\lim_{x\to 1}\left(x^{\frac 1{x-1}}\right)^{\frac {1}{x+1}}\\ &=\lim_{|u|\to \infty}\left(\left(1+\frac 1u\right)^u\right)^{\frac {1}{2+\frac 1u}}\\ &=\sqrt e\thinspace .\end{align} $$

Because, you know that

$$ \begin{align}\lim_{u\to +\infty}\left(1+\frac 1u\right)^u&=\lim_{u\to-\infty}\left(1+\frac 1u\right)^u\\ &=e\thinspace .\end{align} $$

Answer 2

using series expansion

let : $$t = x - 1 \implies t \to 0$$ $$\lim_{t\to0}\left({t＋1}\right)^{\frac {1}{(t＋1)^2-1}}＝\lim_{t\to0}(t＋1)^{\frac {1}{t^2＋2t}}＝\lim_{t\to0}e^{\frac {ln(1＋t)}{t^2＋2t}}＝\lim_{t\to0}e^{\frac {t－\frac {t^2}{2}＋O(t^2)}{t^2＋2t}}＝e^\frac{1}{2}$$

using $$ln(1+t)=t－\frac {t^2}{2}＋O(t^2)$$