$\lim_{x \rightarrow \frac{\pi}{2}} \frac{(\frac{\pi}{2}-x)^2}{\cos x}$
1 Answer
We can L'hopital's rule
\[\lim_{x \rightarrow \frac{\pi}{2}} \frac{(\frac{\pi}{2}-x)^2}{\cos x}=\lim_{x \rightarrow \frac{\pi}{2}} \frac{-2(\frac{\pi}{2}-x)}{-\sin x}=\pm \infty.\]
Indeed the limit does not exists.
Erdos
4.4K
Join Matchmaticians Affiliate Marketing
Program to earn up to 50% commission on every question your affiliated users ask or answer.
- 1 Answer
- 85 views
- Pro Bono
Related Questions
- Question 1 calculus
- Find the volume of the solid obtained by rotating $y=x^2$ about y-axis, between $x=1$ and $x=2$, using the shell method.
- Find $\lim _{x \rightarrow 0} x^{x}$
- Vector field
- Volume of the solid of revolution
- Improper integral convergence
- You have a piece of 8-inch-wide metal which you are going to make into a gutter by bending up 3 inches on each side
- Calculus and Vector
