Find the limit of $\;\lim\limits_{x\to\infty}\left(\frac{x^2-1}{x^2+1}\right)^\frac{x-1}{x+1}$.

Question

Find limit of $\;\lim\limits_{x\to\infty}\left(\dfrac{x^2-1}{x^2+1}\right)^\frac{x-1}{x+1}$ without using $L'Hopital$

I tried subtracting $1$ for using $lim$ of $e$ but I got $1^\infty$ form and couldn't continue.

Answer 1

hint

Use

$$\frac{x^2-1}{x^2+1}=1-\frac{2}{x^2+1}$$

and

$$\lim_{x\to+\infty}\left(1-\frac{2}{1+x^2}\right)^{\frac{x^2+1}{2}}=\frac 1e$$

Answer 2

l'Hopital's rule doesn't apply the to the determinate form $\left[ 1^1 \right]$, so you shouldn't think of applying it here.

You ask explicitly about $\left[ 1^\infty \right]$. Use the inverse relationship between exponentiation and the natural logarithm together with continuity of the exponential function: $$ \lim_{x \rightarrow \infty} f(x)^{g(x)} = \mathrm{e}^{\lim_{x \rightarrow \infty} g(x) \cdot \ln f(x)} \text{.} $$ Now you have $\mathrm{e}^\left[\infty \cdot 0 \right]$

Of relevance to the problem is \begin{align*} &\lim_{x \rightarrow \infty} \frac{a x^n + b x^{n-1} + \cdots + c}{d x^n + e x^{n-1} + \cdots + f} \\ &\quad = \lim_{x \rightarrow \infty} \frac{x^n}{x^n} \cdot \frac{a + b/x + \cdots + c/x^n}{d + e/x + \cdots + f/x^n} \\ &\quad = \frac{a + 0 + \cdots + 0}{d + 0 + \cdots + 0} \text{.} \end{align*}

So either apply this last observation directly and get the determinate form $\left[ 1^1 \right]$ or use the first to get the determinate form $\mathrm{e}^\left[1 \cdot 0 \right]$.