# Find $\lim_{x\to -\infty} \frac{(x-1)^2}{x+1}$

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.

• The denominator goes to $0$ from the left. – xbh Oct 25 '18 at 15:30
• the numerator is always positive as x tends to -infinity, and the denominator is always negative. 1/0 can be positive or negative, because 0 is neither positive nor negative – Cato Oct 25 '18 at 15:39

Hint: Write $$\frac{x^2\left(1-\frac{1}{x}\right)^2}{x\left(1+\frac{1}{x}\right)}$$

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 $$x \rightarrow -\infty$$.

HINT

We have that

$$\lim_{x\to -\infty} \frac{(x-1)^2}{x+1}=\lim_{x\to -\infty} (x-1)\cdot \frac{x-1}{x+1}$$

Hint

If $$x<-1$$ then $$\frac1x+\frac {1}{x^2}<0.$$

First, I think it's better to divide top and bottom by the maximum power of the bottom. In this case, by $$x$$, to get

$$\lim_{x\to -\infty} \frac{\frac{1}{x}(x-1)^2}{1+\frac{1}{x}} = \lim_{x\to -\infty} \frac{\frac{1}{x}(x^2-2x+1)}{1+\frac{1}{x}}$$

$$\lim_{x\to -\infty} \frac{x-2+\frac{1}{x}}{1+\frac{1}{x}}.$$

Now you can see that the top goes to minus infinity and the bottom goes to $$1$$.

$$\lim_{x \to -\infty} \frac{x^2-2x+1}{x+1} \implies \lim_{x \to -\infty} \frac{1-\frac{2}{x}+\frac{1}{x^2}}{\frac{1}{x}+\frac{1}{x^2}}$$ As you mentioned, the numerator tends to $$1$$. However, notice that the denominator tends to $$0^-$$. $$\vert x\vert > 1 \implies \biggr\vert \frac{1}{x}\biggr\vert > \biggr\vert\frac{1}{x^2}\biggr\vert$$ $$\frac{1}{x}+\frac{1}{x^2} < 0$$ Hence, the denomitor tends to $$0^-$$ ($$0$$ from the negative side). Therefore, the limit is $$-\infty$$.

You are playing with dividing by $$0$$

To avoid that, notice $$\lim_{x\to -\infty} \frac{(x-1)^2}{x+1} = \lim_{x\to -\infty} \frac{x^2-2x +1}{x+1}$$

$$= \lim_{x\to -\infty} x-3+\frac {4}{x+1} = -\infty$$