# lim_(x->oo)(sqrt((x^2-1)/(x^2+1)))^((2x+3)/(x-2))= ?

Apr 15, 2017

${\lim}_{x \rightarrow \infty} {\left(\sqrt{\frac{{x}^{2} - 1}{{x}^{2} + 1}}\right)}^{\frac{2 x + 3}{x - 2}} = 1$

#### Explanation:

${\lim}_{x \rightarrow \infty} {\left(\sqrt{\frac{{x}^{2} - 1}{{x}^{2} + 1}}\right)}^{\frac{2 x + 3}{x - 2}}$

Use exponential function and the natural logarithm:

${\lim}_{x \rightarrow \infty} {e}^{\ln} \left({\left(\sqrt{\frac{{x}^{2} - 1}{{x}^{2} + 1}}\right)}^{\frac{2 x + 3}{x - 2}}\right)$

Use the property $\ln \left({a}^{c}\right) = \left(c\right) \ln \left(a\right)$

${\lim}_{x \rightarrow \infty} {e}^{\left(\frac{2 x + 3}{x - 2}\right) \ln \left(\left(\sqrt{\frac{{x}^{2} - 1}{{x}^{2} + 1}}\right)\right)}$

Do that again for the square root:

${\lim}_{x \rightarrow \infty} {e}^{\frac{1}{2} \left(\frac{2 x + 3}{x - 2}\right) \ln \left(\frac{{x}^{2} - 1}{{x}^{2} + 1}\right)}$

Multiply the $\frac{1}{2}$ into the fraction:

${\lim}_{x \rightarrow \infty} {e}^{\left(\frac{2 x + 3}{2 x - 4}\right) \ln \left(\frac{{x}^{2} - 1}{{x}^{2} + 1}\right)}$

Use of L'Hôpital's rule shows that both $\left(\frac{2 x + 3}{2 x - 4}\right)$ and $\frac{{x}^{2} - 1}{{x}^{2} + 1} \to 1 \text{ as } x \to \infty$

$\frac{\frac{d \left(2 x + 3\right)}{\mathrm{dx}}}{\frac{d \left(2 x - 4\right)}{\mathrm{dx}}} = \frac{2}{2} = 1$

$\frac{\frac{d \left({x}^{2} - 1\right)}{\mathrm{dx}}}{\frac{d \left({x}^{2} - 1\right)}{\mathrm{dx}}} = \frac{2 x}{2 x} = 1$

${e}^{1 \ln \left(1\right)} = 1$

Therefore, the limit of the original expression is 1:

${\lim}_{x \rightarrow \infty} {\left(\sqrt{\frac{{x}^{2} - 1}{{x}^{2} + 1}}\right)}^{\frac{2 x + 3}{x - 2}} = 1$

Apr 15, 2017

$1$

#### Explanation:

${\left(\frac{{x}^{2} - 1}{{x}^{2} + 1}\right)}^{\frac{1}{2} \frac{2 x + 3}{x - 2}} = {\left(\frac{1 - \frac{1}{x} ^ 2}{1 + \frac{1}{x} ^ 2}\right)}^{\frac{1}{2} \left(\frac{2 + \frac{3}{x}}{1 - \frac{2}{x}}\right)} = {\left(\frac{1 - \frac{1}{x} ^ 2}{1 + \frac{1}{x} ^ 2}\right)}^{\frac{1 + \frac{3}{2 x}}{1 - \frac{2}{x}}}$

so

${\lim}_{x \to \infty} {\left(\sqrt{\frac{{x}^{2} - 1}{{x}^{2} + 1}}\right)}^{\frac{2 x + 3}{x - 2}} = {\lim}_{x \to \infty} {\left(\frac{1 - \frac{1}{x} ^ 2}{1 + \frac{1}{x} ^ 2}\right)}^{\frac{1 + \frac{3}{2 x}}{1 - \frac{2}{x}}} = {1}^{1} = 1$