Limit of trigonometric function

[needingtoknow]

Homework Statement

lim x --> 0 for function y = (-2x)/(sinx)

Using L'Hopital's Theorem, I found the derivative of the top and of the bottom and found the limit and got -2. How to find the limit as x approaches 0 without using L'Hopital's theorem. I know my solutions manual uses a different method, but I don't know what it is because they have skipped all those steps.

 

[Chestermiller]

Homework Statement

lim x --> 0 for function y = (-2x)/(sinx)

Using L'Hopital's Theorem, I found the derivative of the top and of the bottom and found the limit and got -2. How to find the limit as x approaches 0 without using L'Hopital's theorem. I know my solutions manual uses a different method, but I don't know what it is because they have skipped all those steps.

If they skipped all the steps, how do you know that they did it differently from you? Is it possible that they expanded the denominator in a Taylor series about x = 0?

Chet

 

[jbunniii]

No matter how you solve this problem, you will effectively need to establish that
$$\lim_{x \rightarrow 0}\frac{\sin(x)}{x} = 1$$
There are many ways to do this, L'Hospital's rule being the simplest but not the most elementary.

If you know about Taylor series, then you can expand $\sin(x)$, divide each term by $x$, and take the limit. All terms except the first will go to zero.

There's also nice proof which uses a geometric (not 100% rigorous, but very intuitively convincing) argument to establish the following inequality, which is valid for all nonzero $x \in [-\pi/2, \pi/2]$:
$$\cos(x) \leq \frac{\sin(x)}{x} \leq 1$$
The result then follows from the squeeze theorem. See the first answer at this StackExchange link to see the picture from which the inequality is inferred:

http://math.stackexchange.com/questions/75130/how-to-prove-that-lim-limits-x-to0-frac-sin-xx-1

 

[shortbus_bully]

The simplest way (the first week of Calc 1) to solve this problem is to note that (-2x)/sin(x) =-2/(sin(x)/x) then solve the problem using standard limit rules.

 

[needingtoknow]

Thank you all for posting. I ended up using shortbus_bully's method because he was right it was the simplest way to solve the problem!