I am trying to find the following 2 limits as part of a series of 4 exercises following out lectures. We haven't really learned derivatives yet, so I can't just slap L'Hospital and be done with it.

The first two I could solve with Taylor Expansions, but I still need to solve:

$$\lim_{x\to 0} \frac{e^x -\sin x -1}{x^2}$$ Taylor Expansion of sin didn't work here, and I don't know how to deal with the $e^x$.

$$\lim_{x\to 0} \frac{x\sin x}{\cos x -1}$$ Given $-\sin^2 x = \cos^2 x -1$ I tried multiplying by $\frac{\cos x +1}{\cos x +1}$, which lead me to $\frac{-x\cos x \ -x}{\sin x}$ and then I don't know what else to do. (SOLVED) I suppose I could simplify this to $-x\cot x - \frac{x}{\sin x}$, which leads to -2.


You have




cause $x\mapsto \sin(x)$ is odd.

then the limit is $$\frac{1}{2}$$

| cite | improve this answer | |
  • 1
    $\begingroup$ Thanks, I had completely forgotten I could expand $e^x$ $\endgroup$ – Lucas Vienna Dec 14 '16 at 23:13
  • $\begingroup$ I am trying to solve a similar limit... Can we calculate the above limit only with L'Hospital or with the Taylor expansion? $\endgroup$ – Mary Star Dec 15 '16 at 12:38
  • 1
    $\begingroup$ @MaryStar what he did was the Taylor expansion, you get something like this: $$\frac{(1+x+\frac{x^2}{2!}(1+\epsilon_1(x))\ -(x+x^2\epsilon_2(x))-1}{x^2}$$ $\endgroup$ – Lucas Vienna Dec 15 '16 at 21:16
  • 1
    $\begingroup$ Applying L'Hospital twice would also solve this equation in a pretty easy manner. But then you need to prove a few things, like $\frac{dx}{dy}\sin x = \cos x$ $\endgroup$ – Lucas Vienna Dec 15 '16 at 21:17
  • $\begingroup$ Ah ok... Thank you!! :-) $\endgroup$ – Mary Star Dec 15 '16 at 22:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.