Limit as $x$ tends to infinity

Question

When $\displaystyle f(x) = 1-\frac{1}{x^2}$, then shouldn't the limit as $x \rightarrow -\infty$ and $x \rightarrow +\infty$ be the same? In either case $x^2$ becomes a large positive number. So $1/($a large positive number$)$ tends to $0$, and therefore, the result tends to $1$?

Answer 1

Yes, both limits equal $1$ because $$ \lim_{x\to \infty} \left[1-\frac{1}{x^2}\right]=1-0=1 $$ And $$ \lim_{x\to -\infty} \left[1-\frac{1}{x^2}\right]=1-0=1 $$