Question

Evaluate the limit of (sinx-1)/(cos^2 times x) as x approaches pi over 2.

Answer 1

so um have you done l'hopitals rule ?

Answer 2

In the denominator, do you mean \(\cos^2x\)? The way you've written it sounds like \(x\cos^2x\).
\[\begin{align*}\lim_{x\to\pi/2}\frac{\sin x-1}{\cos^2x}&=\lim_{x\to\pi/2}\frac{\sin x-1}{1-\sin^2x}\\
&=\lim_{x\to\pi/2}\frac{\sin x-1}{(1-\sin x)(1+\sin x)}\\
&=\lim_{x\to\pi/2}\frac{-(1-\sin x)}{(1-\sin x)(1+\sin x)}\\
&=\lim_{x\to\pi/2}-\frac{1}{1+\sin x}
\end{align*}\]
No L'hopital's rule required, but more work is required with this method.

Answer 3

yes thats why i asked if they have covered it in the material because makes things easier