# Trouble with trigonometric limits without derivatives

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

$$e^x=1+x+\frac{x^2}{2}(1+\epsilon_1(x))$$

and

$$\sin(x)=x+x^2\epsilon_2(x)$$

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

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

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