Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

share|improve this question
2  
The restriction sounds so arbitrary to me... –  leonbloy Jan 24 '13 at 16:46
6  
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
1  
Why do you want to compute the limit without L'hopital's Rule or Taylor Series? –  Chris's sis 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

4 Answers 4

up vote 4 down vote accepted

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:

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

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

share|improve this answer
    
+1 I like it very much. –  Jp McCarthy Jan 25 '13 at 13:31
    
@Jp Thank you :) –  L. F. Jan 25 '13 at 15:45
    
Nice solution L. F...... –  juantheron Nov 24 '13 at 18:26

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

share|improve this answer
2  
"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
1  
Take $\varepsilon=1/10$ and $x$ is nowhere near $0$... –  Jp McCarthy Jan 24 '13 at 18:24

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.

share|improve this answer

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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