How to solve this limit? (2)

Question

How to solve this limit?

$$\lim_{x\to\ 0} \frac{x^2+x+1}{x^2-x}$$

Answer 1

$\lim_{x\to\ 0} \frac{x^2+x+1}{x^2-x}=$

$\lim_{x\to\ 0} \frac{(x^2-x)+(2x+1)}{x^2-x}=$

$\lim_{x\to\ 0} 1 + \frac{2x+1}{x^2-x}=$

$1 + \lim_{x\to\ 0} (\frac {2x}{x^2 - x} + \frac 1{x^2 - x})=$

$1 + \lim_{x\to\ 0} (\frac {2}{x-1} + \frac 1{x^2 - x})=$

$1 - 2 + \lim_{x\to\ 0}\frac 1{x^2 -x}$

$-1 + \lim_{x\to\ 0}\frac 1{x(x-1)}$

If $x < 0$ then $x, x-1 < 0$ and this diverges to positive infinity. If $x > 0$ then $x > 0; x-1 < 0$ and this diverges to negative infinity. So there is no reasonable limit.

Answer 2

Hint:

There is no need to use l'Hôpital, since your limit is on the form $\frac{1}{0}$ (and not $\frac{0}{0}$ nor $\frac{\pm\infty}{\infty}$), which either becomes $+\infty$ or $-\infty$ (or neither). Look at limit from two different cases, one where $x>0$ and the other where $x<0$, that is, find:

• $$\lim_{x\rightarrow\ 0^+}\frac{x^2+x+1}{x^2-x}\tag{1}$$
• $$\lim_{x\rightarrow\ 0^-}\frac{x^2+x+1}{x^2-x}\tag{2}$$

Answer 3

On simplification

$$\lim_{x\to\ 0} \frac{x^2+x+1}{x^2-x}\\ = \lim_{x\to\ 0} \frac{(x-1)(x^2+x+1)}{x(x-1)^2}\\ = \lim_{x\to\ 0} \frac{x^3-1}{x(x-1)^2}\\ = \lim_{x\to\ 0} \left(\frac{x^2}{(x-1)^2}- \frac{1}{x(x-1)^2}\right)\\= 0-\lim_{x\to\ 0} \frac{1}{x(x-1)^2}\\=-\lim_{x\to\ 0} \left(\frac{1}{x}-\frac{x-2}{(x-1)^2}\right)\\\ =-2-\lim_{x\to\ 0}\frac{1}{x}$$

I'm guessing you can make your deductions now