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

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.

• Are you allowed to use the identity $\lim_{x\to\infty}\frac1x=0$? Commented Mar 1, 2021 at 14:01
• Yes I am allowed.
– user825688
Commented Mar 1, 2021 at 14:02
• Take my above comment as a hint then, and manipulate both the base and the exponent of the power. This should give you an idea as to what the answer should be. Commented Mar 1, 2021 at 14:03
• I think there is mistake . your limit is simply $1^1=1$ Commented Mar 1, 2021 at 14:09

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 $$\mathrm{e}^\left[1 \cdot 0 \right]$$.

• Good answer ($+1$) Commented Mar 1, 2021 at 19:32

hint

Use

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

and

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