$\lim_{x \rightarrow 0}\left(\frac{1}{\sin^2(x)} -\frac{1}{x^2}\right)$ and the problem of $\infty-\infty$

Asked Sep 10 '16 at 12:38

Active Sep 17 '16 at 16:10

Viewed 86 times

I'm trying to calculate

$$\lim_{x \rightarrow 0}\left(\frac{1}{\sin^2(x)} -\frac{1}{x^2}\right)$$

but I'm having the problem of getting

$$\infty - \infty$$

which I read to be indeterminate.

I cannot figure out how could I manipulate the above to get rid of the indeterminacy.

edited Sep 17 '16 at 16:10

Martin Sleziak

asked Sep 10 '16 at 12:38

mavavilj

2

Write it $$\frac{x^2 - \sin^2 x}{x^2\sin^2 x}$$ and Taylor-expand the numerator. – Daniel Fischer Sep 10 '16 at 12:40
3

See relevant posts: http://math.stackexchange.com/questions/1338411/why-doesnt-using-the-approximation-sin-x-approx-x-near-0-work-for-computin – Wei Zhong Sep 10 '16 at 12:46
1

And http://math.stackexchange.com/questions/629257/evaluate-mathop-lim-limits-x-to-0-left-1-over-sin-2x-1 (searched from https://approach0.xyz/search/?q=%24%5Clim_%7Bx%5Crightarrow0%7D(%5Cfrac%7B1%7D%7B%5Csin%5E2x%7D-%5Cfrac%7B1%7D%7Bx%5E2%7D)%24&p=1) – Wei Zhong Sep 10 '16 at 12:47
What the other answers don't have: This limit is a lot easier if you use the trig identity $\sin^x = (1-\cos 2x)/2.$ – B. Goddard Sep 10 '16 at 16:06

0 Answers0