$\lim_{x \rightarrow \frac{\pi}{2}} \frac{(\frac{\pi}{2}-x)^2}{\cos x}$
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
- 383 views
- Pro Bono
Related Questions
- What is the integral of (x^2-8)/(x+3)dx
- Equations of Motion and Partial Fractions
- Relating integrals to the area under the curve using rectangles.
- Let $ X = x i+ y j+z k$, and $r=||X||$. Prove that $\nabla (\frac{1}{r})=-\frac{X}{r^3}.$
- 2 questions in 1
- Use Rouche’s Theorem to show that all roots of $z ^6 + (1 + i)z + 1 = 0$ lines inside the annulus $ \frac{1}{2} \leq |z| \leq \frac{5}{4}$
- Calc 3 Question
- Calculus - 2nd order differential equations and partial derivatives