Evaluating limit at infinity (algebric issue) Q: evaluate $\lim_{x \to \infty}$ $ (x-1)\over \sqrt {2x^2-1}$
What I did:
when $\lim_ {x \to \infty}$ you must put the argument in the form of $1/x$ so in that way you know that is equal to $0$
but in this ex. the farest that I went was
$\lim_{x \to \infty}$ $x \over x \sqrt{2}$ $1-(1/x) \over (1/x) - ??$
 A: Hint: write $$
\frac{(x-1)}{\sqrt {2x^2-1}} = \frac{x(1-\frac{1}{x})}{x\sqrt{2-\frac{1}{x^2}}}= \frac{1-\frac{1}{x}}{\sqrt{2-\frac{1}{x^2}}}
$$
for $x > 0$. What happens to numerator and denominator when $x\to\infty$?
A: $\frac{x-1}{\sqrt{2x^{2}-1}}=\frac{x-1}{x \sqrt{2-\frac{1}{x^{2}}}}=\frac{1-\frac{1}{x}}{\sqrt{2-\frac{1}{x^{2}}}}$. Hence the limit is $\frac{1}{\sqrt{2}}$
A: $$\lim_\limits{x\to\infty}\dfrac{x-1}{\sqrt{2x^2}}\le
\lim_\limits{x\to\infty}\dfrac{x-1}{\sqrt{2x^2-1}}\le
\lim_\limits{x\to\infty}\dfrac{x-1}{\sqrt{2x^2-2}}$$
$$\lim_\limits{x\to\infty}\dfrac{x-1}{\sqrt{2}x}\le
\lim_\limits{x\to\infty}\dfrac{x-1}{\sqrt{2x^2-1}}\le
\lim_\limits{x\to\infty}\dfrac{x-1}{\sqrt{2(x^2-1)}}$$
$$\lim_\limits{x\to\infty}\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{2}x}\le
\lim_\limits{x\to\infty}\dfrac{x-1}{\sqrt{2x^2-1}}\le
\lim_\limits{x\to\infty}\sqrt{\dfrac{x-1}{2(x+1)}}$$
$$\dfrac{1}{\sqrt{2}}\le
\lim_\limits{x\to\infty}\dfrac{x-1}{\sqrt{2x^2-1}}\le
\lim_\limits{x\to\infty}\sqrt{\dfrac{x+1-2}{2(x+1)}}$$
$$\dfrac{1}{\sqrt{2}}\le
\lim_\limits{x\to\infty}\dfrac{x-1}{\sqrt{2x^2-1}}\le
\lim_\limits{x\to\infty}\sqrt{\dfrac{1}{2}-\dfrac{1}{x+1}}$$
$$\dfrac{1}{\sqrt{2}}\le
\lim_\limits{x\to\infty}\dfrac{x-1}{\sqrt{2x^2-1}}\le
\dfrac{1}{\sqrt{2}}$$
In fact we can say $\lim_\limits{x\to\infty}\dfrac{x-1}{\sqrt{2x^2-1}}\to
\dfrac{1}{\sqrt{2}}$ from beneath.
