I don’t understand something about L’Hôpital’s rule. In this case:

=lim

Why do we have to keep on solving after this step:

\displaystyle\lim_{x\to0}\dfrac{(e^x-2)’}{(12x^2+6x+2)’}

Can’t I just plug in x=0 and compute the limit at this step giving me:

\dfrac{1-2}{0+0+2}=-\dfrac{1}{2}

I’m very confused, because I get different probable answers for the limit, depending on when do I stop to differentiate, as clearly -\frac1{2}\neq \frac 1{24}.

**Answer**

Once your answer is no longer in the form 0/0 or \frac{\infty}{\infty} you must stop applying the rule. You only apply the rule to attempt to get rid of the indeterminate forms. If you apply L’Hopital’s rule when it is not applicable (i.e., when your function no longer yields an indeterminate value of 0/0 or \frac{\infty}{\infty}) you will most likely get the wrong answer.

You should have stopped differentiating the top and bottom once you got to this:

\dfrac{e^x-2x}{4x^2+3x^2+2x}. Taking the limit at that gives you 1/0. The limit is nonexistent.

Also, don’t be tempted to say “infinity” when you see a 0 in the denominator and a non-zero number in the top. It may not be the case. For example, the function \frac{1}{x} approaches infinity and negative infinity from both sides of the limit as x approaches 0. Its not necessarily infinite; its best just to leave it as “nonexistent”.

**Attribution***Source : Link , Question Author : Stella Dimitrova , Answer Author : Michael Hardy*