$\lim_{x \rightarrow \frac{\pi}{2}} \frac{(\frac{\pi}{2}-x)^2}{\cos x}$
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
4.8K
Join Matchmaticians Affiliate Marketing
Program to earn up to a 50% commission on every question that your affiliated users ask or answer.
- 1 Answer
- 359 views
- Pro Bono
Related Questions
- Find limit
- Are my answers correct?
- Rewrite $\int_{\sqrt2}^{2\sqrt2} \int_{-\pi/2}^{-\pi/4} r^2cos(\theta)d\theta dr$ in cartesian coordinates (x,y)
- Find the antiderrivative of $\int \frac{v^2-v_o^2}{2\frac{K_e\frac{q_1q_2}{r^2}}{m} } dr$
- Reduction formulae
- Derivative of $\int_{\sin x}^{x^2} \cos (t)dt$
- Calc 3 Question
- Is $\sum_{i=1}^{\infty}\arctan (\frac{n+1}{n^2+5})$ convergent or divergent?
