Calculate $\lim_{x \to \infty} x(x-\sqrt{x^{2}-1})$ without using L'Hospital's rule Question:
Calculate 
$$\lim_{x \to \infty} x\left(x-\sqrt{x^{2}-1}\right)$$
without using L'Hospital's rule.
Attempted Answer:
My first attempt centers around getting a meaningful answer from the square root expression.
Factoring out $x^{2}$:
$$\lim_{x \to \infty} x\left(x-x\sqrt{1-\frac{1}{x^{2}}}\right)$$
The second term under the square root becomes 0, the square root expression becomes $\sqrt{1} = 1$, then 
$$\lim_{x \to \infty} x(x-x\sqrt{1}) = \lim_{x \to \infty} x(0) \neq \lim_{x \to \infty} 0 = 0$$
Now, I happen to know that the answer to this is $\frac{1}{2}$, so this answer is wrong. This is also clear from the fact that this is a "$0 \cdot \infty$" expression, so my first attempt obviously do not work.
My second attempt centers around the fact that the only thing close to one-half is the square root that can be thought of as raising something to the power of 0.5 or one-half. This suggests that there is a natural logarithm here somewhere where we can move down the 0.5 and everything else cancels or becomes 1:
$$\lim_{x \to \infty} x\left(x-\sqrt{x^{2}-1}\right) = \lim_{x \to \infty}(x^2-x(x^{2}-1)^{\frac{1}{2}}) = $$
$$\lim_{x \to \infty}e^{\ln{x^2}} - e^{\ln{x(x^{2}-1)^{\frac{1}{2}}}}= \lim_{x \to \infty}e^{\ln{x^2}} - e^{\ln{x}+\ln{(x^{2}-1)^{\frac{1}{2}}}}$$
...but this does not seem to go anywhere.
What are some more productive ways to tackle this question?
 A: Hint: Just use $$\ldots = x\frac{(x-\sqrt{x^2-1})(x+\sqrt{x^2-1})}{x+\sqrt{x^2-1}}=\ldots$$
A: $$x\left(x-\sqrt{x^2-1}\right)=\frac{x\left(x-\sqrt{x^2-1}\right)\left(x+\sqrt{x^2-1}\right)}{\left(x+\sqrt{x^2-1}\right)}$$
$$=\frac{x\left(x^2-\left(\sqrt{x^2-1}\right)^2\right)}{x+\sqrt{x^2-1}}=\frac{x}{x+\sqrt{x^2-1}}=\frac{1}{1+\sqrt{1-\frac{1}{x^2}}}\stackrel{x\to \infty}\to\frac{1}{2}$$
A: Here's yet one more approach.  Enforce the substitution $x=\cosh (y)$ and note that as $x\to \infty$, $y\to \infty$.  Then,
$$\begin{align}
\lim_{x\to \infty}x\left(x-\sqrt{x^2-1}\right)&=\lim_{y\to \infty}\cosh (y)\left(\cosh (y)-\sinh (y)\right)\\\\
&=\lim_{y\to \infty}\cosh (y)\,e^{-y}\\\\
&=\lim_{y\to \infty}\frac12\left(1+e^{-2y}\right)\\\\
&=\frac12
\end{align}$$
A: You can try computing a limit at $0$, by $x=1/t$, that changes the limit into
$$
\lim_{t\to0^+}\frac{1}{t}\left(\frac{1}{t}-\sqrt{\frac{1}{t^2}-1}\right)
=
\lim_{t\to0^+}\frac{1-\sqrt{1-t^2}}{t^2}
=
\lim_{t\to0^+}\frac{1-(1-t^2)}{t^2(1+\sqrt{1-t^2})}
=
\frac{1}{2}
$$
