$\lim_{x \rightarrow \frac{\pi}{2}} \frac{(\frac{\pi}{2}-x)^2}{\cos x}$
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.
4.8K
Join Matchmaticians Affiliate Marketing
Program to earn up to a 50% commission on every question that your affiliated users ask or answer.
- 1 Answer
- 441 views
- Pro Bono
Related Questions
- Need help with this calculus question please
- Evaluate $\int \sqrt{\tan x} dx$
- Determine values of a,b and c so that f(0)=0 and f(8)=0 and f'(2)= 16
- Minimizing the cost of building a box
- Evaluate $\int_C (2x^3-y^3)dx+(x^3+y^3)dy$, where $C$ is the unit circle.
- Convergence of $\int_{1}^{\infty} e^{\sin(x)}\cdot\frac{\sin(x)}{x^2} $
- Question 1 calculus
- Finding the arc length of a path between two points