limits problems--need help!

Printable View

February 23rd 2010, 05:19 PM
kdl00

limits problems--need help!

I'm stuck on getting the limits for sin/cos problems... mind helping me solve these problems?

1. lim x-->0 x^2/1-cosx
2. lim x-->0 x/sin3x
3. lim x-->0 sin5x/sin2x

(I don't know how to type limit problems in this forum..)
February 23rd 2010, 05:22 PM
icemanfan

Are you allowed to use L'Hospital's Rule? If so,

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

$\lim_{x \to 0} \frac{2x}{\sin x} =$

$\lim_{x \to 0} \frac{2}{\cos x} = 2$

You can use the same approach for the other two problems.
February 23rd 2010, 05:26 PM
Locke

Quote:

Originally Posted by kdl00

I'm stuck on getting the limits for sin/cos problems... mind helping me solve these problems?

1. lim x-->0 x^2/1-cosx
2. lim x-->0 x/sin3x
3. lim x-->0 sin5x/sin2x

(I don't know how to type limit problems in this forum..)

Neither do I. But I can help you. :)

Use L'hopital's Rule. If you have a limit that approaches 0, the function is a fraction, and the solution comes out undefined, get the derivative of the numerator, and the derivative of the denominator, and you should be able to solve it. I'll do number one:

lim x-->0 x^2/1-cosx

becomes:

lim x-->0 2x/sinx

becomes:

lim x-->0 2/cosx

becomes:

2/1

and you can repeat it as much as you like as long as the limit approaches zero.
February 23rd 2010, 05:37 PM
mr fantastic

Quote:

Originally Posted by kdl00

I'm stuck on getting the limits for sin/cos problems... mind helping me solve these problems?

1. lim x-->0 x^2/1-cosx
2. lim x-->0 x/sin3x
3. lim x-->0 sin5x/sin2x

(I don't know how to type limit problems in this forum..)

1. Either use l'Hopital's Rule or substitute the Maclaurin series for cos x and simplify. Or use a useful trig limit found here: World Web Math: Useful Trig Limits

2. Note that it can be writen as $\left(\frac{1}{3}\right) \frac{(3x)}{\sin (3x)}$ . Now use a useful trig limit.

3. Note that it can be written as $\left( \frac{5}{2}\right) \frac{\sin (5x)}{(5x)} \cdot \frac{(2x)}{\sin (2x)}$ . Now use a useful trig limit.