I was wondering if anyone could help me with computing a limit using L'Hôpital's Rule.

Using L'Hôpital Rule for the following limit, I get the following result:

\begin{equation} \lim_{x \to 0} \frac{e^x-1-x^2}{x^4+x^3+x^2} \therefore \lim_{x \to 0} \frac{e^x-1-x^2}{x^4+x^3+x^2} \stackrel{L'H}{=} \lim_{x \to 0} \frac{e^x-2x}{4x^3+3x^2+2x} \end{equation}

(1/0) which means that the limit is either $+\infty$, $-\infty$ or it does not exist.

If I want to find out the answer, I should take the side limits - but here lies my question: the side mimits of what? The original function or the function after L'Hôpital rule? Or any of the two?

Should I do this

\begin{equation} \lim_{x \to 0^-} \frac{e^x-2x}{4x^3+3x^2+2x} = -\infty \end{equation} \begin{equation} \lim_{x \to 0^+} \frac{e^x-2x}{4x^3+3x^2+2x} = +\infty \end{equation}

Or this?

\begin{equation} \lim_{x \to 0^-} \frac{e^x-1-x^2}{x^4+x^3+x^2} = -\infty \end{equation} \begin{equation} \lim_{x \to 0^+} \frac{e^x-1-x^2}{x^4+x^3+x^2} = +\infty \end{equation}

Coincidentally (or not) both limits (from the right of both equations and from the left of both equations) give the same answer. Therefore, the limit does not exist.

For the other function below, the same thing happens:

\begin{equation} \lim_{x \to 0} \frac{sin(x)}{x^4+x^3} \end{equation}

So what I dind't find out is which function I should use, or if it really doens't matter.

Thank you


If you have the possibility you can always make a plot (for example with wolfram alpha) to check if your calculations are correct. I come to the same result for the first limit however for the second one I get $\lim_{x\to 0}\frac{\sin(x)}{x^4+x^3}=\lim_{x\to 0}\frac{\cos(x)}{4x^3+x^2}$, which tends to $+\inf$ for both limits (notice that $x^2$ dominates $x^3$ in this case)

  • $\begingroup$ Hi user159543, thanks for your answer! You are completely right - the second limit is completely different: using the original function we get $+\infty$ and $-\infty$ (so the limit doesn't exist) and using the function after L'Hôpital's Rule we get $+\infty$ and also $+\infty$ (so $+\infty$). So the answer is nonexistent correct? Best Regards. $\endgroup$ – bru1987 Jun 26 '14 at 19:14
  • $\begingroup$ the orginal function behaves like $\frac{\sin(x)}{x^3}$ close to $0$ which behaves like $\frac{x}{x^3}=\frac{1}{x^2}$ so $+\infty$ for both limits $\endgroup$ – Michael Jun 26 '14 at 19:24
  • $\begingroup$ Hi Michael,you are right, both limits are $+\infty$ for the function with sin(x). Best Regards. $\endgroup$ – bru1987 Jun 26 '14 at 19:37
  • $\begingroup$ happy we didn't break maths :) if my answer was useful to you, you can mark it as such with the arrow symbol on the left. $\endgroup$ – Michael Jun 26 '14 at 19:40
  • $\begingroup$ Michael I wish I could but I must have 15 points to do that (this was my first interaction with the website). I did however give you a green mark, I hope that counts for your reputation! Have a great day, Best Regards. $\endgroup$ – bru1987 Jun 26 '14 at 19:52

The right way to proceed is the following from what you have found

$$\lim_{x \to 0^+} \frac{e^x-2x}{4x^3+3x^2+2x} = \infty \implies \lim_{x \to 0^+} \frac{e^x-1-x^2}{x^4+x^3+x^2} = \infty$$

$$\lim_{x \to 0^-} \frac{e^x-2x}{4x^3+3x^2+2x} = -\infty \implies \lim_{x \to 0^-} \frac{e^x-1-x^2}{x^4+x^3+x^2} = -\infty$$

where the LHS are evaluated by l'Hospital's rule and the RHS are derived form this result.

Therefore in this case we conclude that limit doesn't exist.

For the other one we can also proceed as follows

$$\frac{\sin(x)}{x^4+x^3}=\frac{\sin(x)}{x}\frac{x}{x^4+x^3}=\frac{\sin(x)}{x}\frac{1}{x^3+x^2} \to 1 \cdot \infty=\infty$$


L'Hôpital's Rule is applicable when say f and g are are differentiable functions on open intervals K except at maybe a point 'c'.

Mathematically put, IF $\lim_{x\to c} f(x) = \lim_{x\to c} g(x) = 0 $ or $\pm\infty$, and $ g'(x) \neq 0$ for all x in K except $x\neq c$

THEN $\lim_{x \to c} \frac{f(x)}{g(x)}=\lim_{x \to c} \frac{f'(x)}{g'(x)}$.

In your equation, thus L'Hôpital's Rule can be applied repeatedly to get the answer 1/24.

  • $\begingroup$ You are very incorrect. $\endgroup$ – amWhy Jun 26 '14 at 19:29
  • $\begingroup$ L'Hôpital could only be used if the expression is indeterminate so you could only aply the rule once, since $e^x-2x\to 1$ $\endgroup$ – Michael Jun 26 '14 at 19:30
  • $\begingroup$ Hi Anjali, thanks for your answer. I agree with amWhy, you can verify that your answer is not right when plotting the function. Best Regards. $\endgroup$ – bru1987 Jun 26 '14 at 19:36

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