# Quick questions on L'Hopital's rule

1. Jan 4, 2015

### mrg

1. The problem statement, all variables and given/known data
$$\lim_{x\rightarrow -2^-}\frac{x}{x^2+x-2}$$

2. The attempt at a solution

Clearly, when graphing the above equation, the limit does not exist (or approaches positive infinity). However, when applying L'Hopital's Rule, we have $$\frac{1}{2x+1}$$ and then we can go ahead and substitute -2 in place of x, which yields a result of -1/3. Can someone kindly explain the discrepancy?

2. Jan 4, 2015

### deskswirl

What makes you think that L'Hopital's rule is applicable?

3. Jan 4, 2015

### mrg

It's a limit of the form $$\lim_{x\rightarrow a} \frac{f(x)}{g(x)}$$.

4. Jan 4, 2015

### RUber

L'hopital's rule requires that both functions are not defined at the point in question. If you have a function that goes to $\frac{c}{0}$ it will surely go to $\pm \infty$.
Using L'hopital's rule when it does not apply will almost always give you the wrong answer.
The reason is that as functions approach infinity or become infinitely small, their behavior will be relative to their derivatives.

5. Jan 4, 2015

### mrg

Got it. Thanks for the help!

6. Jan 4, 2015

### deskswirl

L'Hopital's rule applies in situation where the above ratio is indeterminate. This means evaluating both f(x) and g(x) as x approaches a either both yields zero or infinity. In your case the numerator is just a number (-2) and the denominator is zero. Therefore the ratio is not indeterminate as x approaches -2 from the left (Cannot use L'Hopital).