Question

Limit as x approaches 0 to cos -1 /x^2
How to solve it?

Answer 1

Can you use l'Hôpital's rule?

Answer 2

actually, to be sure, do you mean \[\frac{\cos x -1}{x^2} \]

Answer 3

Look at the denominator x^2...as x gets closer to 0 what happens to that denominator.

Answer 4

\[ \large \lim_{x \rightarrow0} \frac{\cos x - 1}{x^2}, \text{ form 0/0, use l'Hôpital's rule} \\ \, \\
\large \lim_{x \rightarrow0} \frac{-\sin x}{2x}, \text{ form 0/0, use l'Hôpital's rule} \\
\, \\
\large \lim_{x \rightarrow0} \frac{-\cos x}{2}=\frac{-\cos 0}{2}=\frac{-1}{2}\]

Answer 5

Yes, that what i meant kirbykirby....my prof never taught us I'Hopital's rule, is there not any other way to get to the answer?

Answer 6

Oh ok I finally thought of a way :)

Answer 7

Recall the important limit \[ \large \lim_{x\rightarrow 0}\frac{\sin x}{x}=1\\
\]
Multiply the numerator and denominator by \(\cos x + 1\):
\[\large \large \lim_{x\rightarrow 0}\frac{\cos x-1}{x^2}\frac{\cos x +1}{\cos x +1}\\ ~ \\=\large \lim_{x\rightarrow 0}\frac{\cos^2 x-1}{x^2(\cos x +1)}, ~\small\text{using the differene of squares }x^2-a^2=(x+a)(x-a) \\~\\
\large =\lim_{x\rightarrow 0}\frac{-\sin^2x}{x^2(\cos x +1)}, ~\text{ since } \cos^2x+\sin^2x=1\\~\\ \large =\lim_{x\rightarrow 0}-\frac{\sin^2x}{x^2}\frac{1}{\cos x + 1}\\
~ \\ = \large \lim_{x\rightarrow 0}-\left( \frac{\sin x}{x}\right)^2 \frac{1}{\cos x +1}\\
~ \\
= - (1)^2\cdot \frac{1}{2}=\frac{-1}{2}\]