Find $\lim_{x\to -\infty} \frac{(x-1)^2}{x+1}$
If I divide whole expression by maximum power i.e. $x^2$ I get,$$\lim_{x\to -\infty} \frac{(1-\frac1x)^2}{\frac1x+\frac{1}{x^2}}$$
Numerator tends to $1$ ,Denominator tends to $0$
So I get the answer as $+\infty$
But when I plot the graph it tends to $-\infty$
What am I missing here? "Can someone give me the precise steps that I should write in such a case." Thank you very much!
NOTE: I cannot use L'hopital for finding this limit.
Best Answer
The point is that the denominator not just tends to zero, but tends to zero from the left, i.e. from being negative.
Alternatively, rewrite like this:
$$\frac{(x-1)^2}{x+1} = \frac{(x+1)^2-4x}{x+1}=x+1-\frac{4}{1+\frac{1}{x}}$$
which clearly tends to $-\infty$ as $x \rightarrow -\infty$.