Question

I don't understand why the limit of x/sqrt(x^2 - 1) when x -> infinity is 1, and when x -> - infinity is -1
anybody ?

Answer 1

First, as \( x \rightarrow -\infty\), we know the numerator x is negative
and the denominator \( \sqrt{x^2 -1}\) is positive (we take the principal i.e. positive root)
thus the limit must be negative.

on the other hand, for \( x \rightarrow +\infty\), the limit will be positive.

Answer 2

to find the numerical value of the limit, we divide top and bottom by x to get
\[ \frac{1}{\frac{\sqrt{x^2-1}}{x}} \]
for \(x \rightarrow +\infty\) we can replace x with \( \sqrt{x^2}\) (again we assume the principal root) and we have
\[ \frac{1}{\sqrt{\frac{x^2-1}{x^2}}}= \frac{1}{\sqrt{1-\frac{1}{x^2}}}\]
and as x approaches infinity 1/x^2 goes to zero, and we have
\[ \lim_{x \rightarrow \infty} \frac{1}{\sqrt{1-\frac{1}{x^2}}}= \frac{1}{\sqrt{1}}= 1\]

for \(x \rightarrow -\infty\) , we do the same thing, but we must write
\[x = - \sqrt{x^2} \] to get the correct sign.
\[ \lim_{x \rightarrow -\infty} \frac{1}{\frac{\sqrt{x^2-1}}{-\sqrt{x^2}}} \\
= -\frac{1}{\sqrt{1-\frac{1}{x^2}}}= -1
\]