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Limit is tending to infinity (remember without L'Hospital's Rule)

$$\lim_{x \to \infty} x\left[\sqrt{x^2+1}-\sqrt{x^2-1}\right]^2$$

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    L'hospital is not for that kind of case I think. The rule is for $p(x)/q(x)$ type where $p$ and $q$ both converge or to zero or to infinity – 2017-02-24
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    @RafaelWagner I want the solution without L'Hospital's Rule – 2017-02-24
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    $\sqrt{x^2+1}-\sqrt{x^2-1}=\frac{2}{\sqrt{x^2+1}+\sqrt{x^2-1}}$ – 2017-02-24

2 Answers 2

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$$\lim_{x\to\infty} x\left[\sqrt{x^2+1} - \sqrt{x^2-1}\right]^2$$

$\begin{array}\\ \sqrt{x^2+1} - \sqrt{x^2-1} &=(\sqrt{x^2+1} - \sqrt{x^2-1})\dfrac{\sqrt{x^2+1} + \sqrt{x^2-1}}{\sqrt{x^2+1} + \sqrt{x^2-1}}\\ &=\dfrac{x^2+1 - (x^2-1)}{\sqrt{x^2+1} + \sqrt{x^2-1}}\\ &=\dfrac{2}{\sqrt{x^2+1} + \sqrt{x^2-1}}\\ \text{so}\\ x\left[\sqrt{x^2+1} - \sqrt{x^2-1}\right]^2 &=x\left(\dfrac{2}{\sqrt{x^2+1} + \sqrt{x^2-1}}\right)^2\\ &=\dfrac{4x}{(\sqrt{x^2+1} + \sqrt{x^2-1})^2}\\ &=\dfrac{4x}{(x(\sqrt{1+1/x^2} + \sqrt{1-1/x^2}))^2}\\ &=\dfrac{4}{x(\sqrt{1+1/x^2} + \sqrt{1-1/x^2})^2}\\ &\to 0\\ \end{array} $

If it had $x^2\left[\sqrt{x^2+1} - \sqrt{x^2-1}\right]^2 $, the limit would be $1$.

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HINT Expand the square to get $$ (x^2+1) + (x^2-1) - 2\sqrt{(x^2+1)(x^2-1)} = 2x^2 - 2 \sqrt{x^4-1} $$ and now expand the root into Taylor series.

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    Instead of Taylor series, you could also multiplicate by the "conjugate" ... – 2017-02-24