lim x-> 0 of f(x) = (4/x^2 - 2/(1-cosx))

[tdotgirl]

How do I evaluate this limit using l'Hopital's rule?

lim x-> 0 of f(x) = (4/x^2 - 2/(1-cosx))

After a few steps I get stuck.

 

[royhaas]

If you use L'Hospital's Rule directly, you will need to apply it several times. An alternative is to expand in a Maclaurin series, but you must use enough terms. In this case, use 1-cos(x) = x^2/2 - x^4/24. You can also apply L'H once, then use the expansion sin(x) = x-x^3/6 as well. In other words, you need one more term than the highest polynomial, which is x^2, in the expansion. When you simplify and cancel powers of x, you can take the limit of what's left. You should get -1/3 when you're done.

 

[tdotgirl]

ok, i'll try that. thanks!

 

[galactus]

We can use L'Hopital to do this, but like roy said, it must be applied several times.

Rewrite \(\displaystyle \L\\\frac{4}{x^{2}}-\frac{2}{1-cos(x)}=\frac{-2(-2+2cos(x)+x^{2})}{x^{2}(1-cos(x))}\)

Now use ol' L'Hopital and we get:

\(\displaystyle \L\\-4\lim_{x\to\0}\frac{x-sin(x)}{x(2-2cos(x)+xsin(x))}\)

Again:

\(\displaystyle \L\\4\lim_{x\to\0}\frac{cos(x)-1}{2-2cos(x)+4xsin(x)+xsin(x)}\)

Again:

\(\displaystyle \L\\4\lim_{x\to\0}\frac{sin(x)}{-6sin(x)-6xcos(x)+x^{2}sin(x)}\)

Again:

\(\displaystyle \L\\4\lim_{x\to\0}\frac{cos(x)}{-12cos(x)+8xsin(x)+x^{2}sin(x)}\)

\(\displaystyle \L\\4\left[\frac{\lim_{x\to\0}cos(x)}{-12\lim_{x\to\0}cos(x)+8\lim_{x\to\0}xsin(x)+\lim_{x\to\0}x^{2}sin(x)}\right]\)

\(\displaystyle \L\\4\left(\frac{1}{-12+0+0}\right)=\frac{-4}{12}=\frac{-1}{3}\)

Therefore, the limit is -1/3