1. ## Limits question

Well I dont get this to add upp:
It is the following:
$\displaystyle \lim_{x \to 0} \frac{x^2}{1-cos^2(x)}$

Here is my work:

$\displaystyle \lim_{x \to 0} \frac{x}{sinx} \times \frac{x}{sinx}$
and since $\displaystyle \frac{x}{sinx}$ is the reciprocal of $\displaystyle \frac{sinx}{x}$
which is one that should be:
$\displaystyle \lim_{x \to 0} \frac{1}{1} \times \frac{1}{1} = 1$
What am I doing wrong, it said solve and few pages back explained some reciprocal method..

clearly I am doing something very bad...asking my prof what would lead to him $\displaystyle \rightarrow$ and me

3. The following fundamental inequality figures prominently below (as it does with many fundamental properties between cos and sin):

For $\displaystyle 0 < x <\frac{\pi}{2},$
$\displaystyle 0 < cos(x) <\frac{sin(x)}{x} < \frac{1}{cos(x)}$

This implies that:

$\displaystyle cos^2(x) < \frac{x^2}{sin^2(x)} < \frac{1}{cos^2(x)}$

As x $\displaystyle \rightarrow$ 0, the squeeze is put on the middle expression, and it evaluates to 1.

4. the sandwich theorem?!

5. your way is also valid but it requires using the limit of the $\displaystyle \frac{sinx}{x}$ result which is first proven using the squeeze theorem (sandwich). In any case, you used some words in your argument that should be replaced with equations:
$\displaystyle \lim_{x \to 0} \frac{x}{sinx} \times \frac{x}{sinx}$ =
$\displaystyle \lim_{x \to 0} \frac{1}{\frac{sinx}{x}} \times \lim_{x \to 0} \frac{1}{\frac{sinx}{x}}$

On-Edit: Random vairable's way is cool also. Btw, one can also use L'hopital here...lol...no shortage of methods.

6. $\displaystyle \lim_{x \to 0} \ \frac{x^{2}}{\sin^{2}x} = \lim_{x \to 0} \ \frac{x}{\sin x} \cdot \lim_{x \to 0} \ \frac{x}{\sin x}$

$\displaystyle \lim_{x \to 0} \ \frac{x}{x - \frac{x^{3}}{3!} +\frac{x^{5}}{5!} - \frac{x^{7}}{7!} + ...} \cdot \lim_{x \to 0} \ \frac{x}{x - \frac{x^{3}}{3!} +\frac{x^{5}}{5!} - \frac{x^{7}}{7!} + ...}$

$\displaystyle = \lim_{x \to 0} \ \frac{1}{1 - \frac{x^{2}}{3!} +\frac{x^{4}}{5!} - \frac{x^{6}}{7!} + ...} \cdot \lim_{x \to 0} \ \frac{1}{1 - \frac{x^{2}}{3!} +\frac{x^{4}}{5!} - \frac{x^{6}}{7!} + ...} = 1 \cdot 1 = 1$

7. $\displaystyle \lim_{x \to 0} \frac{x^2}{1-cos^2(x)}= \lim_{x \to 0}\dfrac{x^2}{\sin^2(x)}=$

$\displaystyle \left( \lim_{x \to 0}\dfrac{x}{\sin(x)}\right)^2 = \left( \dfrac{1}{\lim_{x \to 0}\frac{\sin(x)}{x}}\right)^2=1$

End.

8. Originally Posted by vince
your way is also valid but it requires using another result.i'd frame it like this:
$\displaystyle \lim_{x \to 0} \frac{x}{sinx} \times \frac{x}{sinx}$ =
$\displaystyle \lim_{x \to 0} \frac{1}{\frac{sinx}{x}} \times frac{1}{\frac{sinx}{x}}$
Im sorry som error in the laTex script, anyway

The solution is telling me that the limit is not 1 but 0.
Here is the books "correct solution".

$\displaystyle \lim_{x \to 0} \frac{x^2}{1-cos^2x} = \lim_{x \to 0} \frac{x}{1-cosx} \times \frac {x}{1+cosx}$
and then since the $\displaystyle \frac{x}{1-cosx} \rightarrow \frac{0}{1}$ and
and the $\displaystyle \frac{x}{1+cosx} \rightarrow \frac{0}{1+1}$ then
$\displaystyle \lim_{x \to 0} = \frac{0}{1} \times \frac{0}{1+1} = 0$

9. Originally Posted by Henryt999
Im sorry som error in the laTex script, anyway

The solution is telling me that the limit is not 1 but 0.
Here is the books "correct solution".

$\displaystyle \lim_{x \to 0} \frac{x^2}{1-cos^2x} = \lim_{x \to 0} \frac{x}{1-cosx} \times \frac {x}{1+cosx}$
and then since the $\displaystyle \frac{x}{1-cosx} \rightarrow \frac{0}{1}$ and
and the $\displaystyle \frac{x}{1+cosx} \rightarrow \frac{0}{1+1}$ then
$\displaystyle \lim_{x \to 0} = \frac{0}{1} \times \frac{0}{1+1} = 0$
$\displaystyle \dfrac{x}{1-\cos(x)}\to \dfrac{0}{0}$ Indefinition. You have many correct solutions in this topic, man.

10. ouch...their solution is legit. something sneaky is going on again...jeez..ill get back to you...

On-edit: nice catch Felper...oh yeah...glossed over that one.

11. Originally Posted by felper
$\displaystyle \dfrac{x}{1-\cos(x)}\to \dfrac{0}{0}$ Indefinition. You have many correct solutions in this topic, man.
How does that go to $\displaystyle \frac{0}{0}$
doesn´t it approach one of the infinities when that happens?

12. The right answer is 1. because they use that indeterminate expression in their logic, anything that followed that point is indeterminate. therefore, they should have stopped there, and either applied L'hopital to that indeterminate expression or used other methods such as those shown above by felper (or you), Random variable, or me.