# Thread: Limit Involving Squeeze Theorem

1. ## Limit Involving Squeeze Theorem

The problem I have is $\lim_{{x}\to{0}}\frac{cosx - 1}{x^2}$

I know I have to use the squeeze theorem, by implementing the range of the cosine function; but when I do that I get

$\frac{-1}{x^2}\leq\frac{cosx}{x^2}\leq\frac{1}{x^2}$

But that would imply the original limit would go to approach zero, but that is certainly not what the answer key attests. Am I setting up my range wrong?

2. ## Re: Limit Involving Squeeze Theorem

Do you have to use squeeze theorem? What about L'Opital's rule?

3. ## Re: Limit Involving Squeeze Theorem

Hello Bashyboy,

If we were to try to solve this problem with the squeeze theorem using the bound conditions for $\cos x$, we would have:

$-1 \leq \cos x \leq 1$

$\Rightarrow -2 \leq \cos x -1 \leq 0$

$\Rightarrow \frac{-2}{x^2} \leq \frac{\cos x -1}{x^2} \leq \frac{1}{x^2}$

$\Rightarrow -\infty \leq \lim_{{x}\to{0}} \frac{\cos x -1}{x^2} \leq \infty$

which doesn't really help us at all.

Provided you are allowed to use L'Hospital's Rule, we can say:

$\lim_{{x}\to{a}} \frac{f(x)}{g(x)} = \lim_{{x}\to{a}} \frac{f'(x)}{g'(x)}$

$\frac{d(\cos x -1)}{dx} = -\sin x$

and...

$\frac{d(x^2)}{dx} = 2x$

so...

$\lim_{{x}\to{0}} \frac{\cos x -1}{x^2} = \lim_{{x}\to{0}} \frac{-\sin x}{2x} = -1/2$

If you HAVE to use the squeeze theorem, just play around with functions that are ALWAYS outside of the bounds of $\cos x$ and see what happens. Good luck!

4. ## Re: Limit Involving Squeeze Theorem

Ah, yes, that would be profoundly simple. Though, I wonder, would it be possible to do the squeeze theorem?

5. ## Re: Limit Involving Squeeze Theorem

I'm not really sure if you can do the squeeze theorem here. If it is possible, I definitely can't think of the bound functions at the moment lol.

6. ## Re: Limit Involving Squeeze Theorem

not a squeeze, but not L'Hopital either ...

$\frac{\cos{x}-1}{x^2} \cdot \frac{\cos{x}+1}{\cos{x}+1} =$

$\frac{\cos^2{x}-1}{x^2(\cos{x}+1)} =$

$\frac{-\sin^2{x}}{x^2(\cos{x}+1)}$

$-\frac{\sin{x}}{x} \cdot \frac{\sin{x}}{x} \cdot \frac{1}{\cos{x}+1}$

now take the limit as $x \to 0$ ...

7. ## Re: Limit Involving Squeeze Theorem

How do you find limit of sinx/x without L'hospital? That's not easier than original

Originally Posted by skeeter
not a squeeze, but not L'Hopital either ...

$\frac{\cos{x}-1}{x^2} \cdot \frac{\cos{x}+1}{\cos{x}+1} =$

$\frac{\cos^2{x}-1}{x^2(\cos{x}+1)} =$

$\frac{-\sin^2{x}}{x^2(\cos{x}+1)}$

$-\frac{\sin{x}}{x} \cdot \frac{\sin{x}}{x} \cdot \frac{1}{\cos{x}+1}$

now take the limit as $x \to 0$ ...

8. ## Re: Limit Involving Squeeze Theorem

Originally Posted by httr
How do you find limit of sinx/x without L'hospital? That's not easier than original
at the start of most calculus courses, one learns the geometric proof of

$\lim_{x \to 0} \frac{\sin{x}}{x} = 1$

this basic trig limit is required in finding the derivative of $y = \sin{x}$ using first principles ...

$f'(x) = \lim_{h \to 0} \frac{\sin(x+h) - \sin{x}}{h}$