# Find limit without using L'hopital or Taylor's series

I'm trying to solve this limit $without$ using L'hopital's Rule or Taylor Series. Any help is appreciated!

$$\lim\limits_{x\rightarrow 0^+}{\dfrac{e^x-\sin x-1}{x^2}}$$

• The restriction sounds so arbitrary to me... – leonbloy Jan 24 '13 at 16:46
• Sounds good. Find the local behaviour of the function without looking at the local behaviour of the function. – André Nicolas Jan 24 '13 at 16:53
• @user59518: Are you basically asking to do the limit using first principles from the definition? Regards – Amzoti Jan 24 '13 at 16:59
• Why do you want to compute the limit without L'hopital's Rule or Taylor Series? – user 1357113 Jan 24 '13 at 17:06
• @Chris'ssister: I agree with you. I find no reason to avoid this two method and make the problem harder. – A.D Jan 24 '13 at 17:09

One possible way is to shoot linear functions at the limit - not very elegant, but it works. Let:

$$f(x)=x^3-\frac{x^2}{2}+e^x-\sin x-1,\;\;x\geq 0$$

Computing the first few derivatives of $f:$

$$f'(x)=3x^2-x+e^x-\cos x$$ $$f''(x)=6x-1+e^x+\sin x$$

$f''$ is clearly increasing and since $f''(0)=0$ we have $f''(x)>0$ for $x\in (0,a)$ for some $a$. This in turn implies that $f'$ is strictly increasing and since $f'(0)=0$ we again have $f'(x)>0$ for $x\in (0,a)$. Finally, this means $f$ is also increasing on this interval, and since $f(0)=0$ we have:

$$0\leq x\leq a:\quad f(x)\geq 0$$

$$\Rightarrow \;\;\frac{e^x-\sin x-1}{x^2}\geq \frac{1}{2}-x$$

Similarly by considering $h(x)=-x^3-\dfrac{x^2}{2}+e^x-\sin x-1$ it is very easy to show that:

$$0\leq x\leq b: \quad h(x)\leq 0$$

$$\Rightarrow \;\;\frac{1}{2}+x\geq\frac{e^x-\sin x-1}{x^2}$$

Hence for small positive $x$ we have:

$$\frac{1}{2}-x\leq\frac{e^x-\sin x-1}{x^2}\leq \frac{1}{2}+x$$

$$\lim_{x\to 0^+}\frac{e^x-\sin x-1}{x^2}=\frac{1}{2}$$

I have an idea that might be fruitful although I haven't been very careful and completely ignore the $x\rightarrow 0^+$! Define the complex function

$\displaystyle F(z)=\frac{e^z-\sin z-1}{z^2}$.

Now let $z=it$.

$\displaystyle F(z)=\frac{e^{it}-\sin(ti)-1}{t^2}=\frac{\cos t+i\sin t-\sin(ti)-1}{t^2}$

$\displaystyle \Rightarrow F(z)=\frac{\cos t-1}{t^2}+\frac{i\sin t-\sin(it)}{t^2}$.

Now take the limit as $t\rightarrow0$. It is not difficult to show that the first limit is one half (multiply above and below by $\cos t+1$). I do not know how to compute the second limit which is annoying me although I think that it is zero.

Now use the fact that

$\displaystyle \lim_{x\rightarrow 0}\,F(x)=\lim_{z\rightarrow 0}\,F(z)=\lim_{t\rightarrow 0}\,F(it)$

I have assumed that the middle limit exists which is troublesome.

$$\lim_{x \to 0^+}\frac{e^x-\sin x -1}{x^2}=\lim_{x \to 0^+}\frac{e^x- x -1}{x^2}+\lim_{x \to 0^+}\frac{x-\sin x}{x^2}$$ First part given here and next part $$|\frac{x -\sin x}{x^2}-0| < \frac{x}{x^2}< \epsilon \implies x >\frac{1}{\epsilon}$$ Take $N(\epsilon)= \lfloor \frac{1}{\epsilon}\rfloor+1$ So here from definition of limit we get $$\lim_{x \to 0^+}\frac{x-\sin x}{x^2} =0$$ Otherwise see this

• "without using Taylor series" – leonbloy Jan 24 '13 at 16:43
• -1. OP: "without using L'hopital's Rule or Taylor Series" – dwarandae Jan 24 '13 at 16:43
• Although the correct answer pops out I think your evaluation of the second limit is incorrect. The limit is additive when the constituent limits are finite but $\lim_{x\rightarrow 0^+}$ is not. – JP McCarthy Jan 24 '13 at 17:53
• @JpMcCarthy I think you are right . So I use definition of limit. – A.D Jan 24 '13 at 18:22
• Take $\varepsilon=1/10$ and $x$ is nowhere near $0$... – JP McCarthy Jan 24 '13 at 18:24

Perhaps the problem says $x\to 0^+$ because some inequalities wich relate $\sin x$ and $e^x$ with polynomials are only valid for $x>0$ sufficiently close to $0$. So, a strategy is to find functions $f(x)$ and $g(x)$ such that

$$f(x)\leq \dfrac{e^x-\sin x-1}{x^2}\leq g(x)\:\;\mbox{ for }x>0 \mbox{ close to } 0$$

and $\displaystyle\lim_{x\to 0^+}f(x)=\lim_{x\to 0^+}g(x)=\frac{1}{2}$