Question

How do you find the limit of
1−sin^3θ/cos^2θ as theta approaches pi/2?​

Answer 1

SOLUTION:

To Determine: The value of the limit.

Given Limit:

$\displaystyle \rm = \lim_{ \theta \to \frac{\pi}{2} } \bigg( \dfrac{1 - \sin^{3} \theta }{ { \cos}^{2} \theta } \bigg)$

If we put θ = π/2, we get:

$\displaystyle \rm = \dfrac{1 - \sin^{3} \frac{\pi}{2} }{ { \cos}^{2} \frac{\pi}{2} }$

$\displaystyle \rm = \dfrac{1 - 1}{0}$

$\displaystyle \rm = \dfrac{0}{0}$

Which is an indeterminate form.

So, let us apply L'Hopital's rule to solve this.

Using L'Hopital's rule, we get:

$\displaystyle \rm = \lim_{ \theta \to \frac{\pi}{2} } \dfrac{ \frac{d}{d \theta}( 1 - \sin^{3} \theta) }{ \frac{d}{d \theta} { \cos}^{2} \theta }$

$\displaystyle \rm = \lim_{ \theta \to \frac{\pi}{2} } \dfrac{ 0 - 3\sin^{2} \theta \cos \theta }{ - 2 \sin \theta \cos\theta }$

$\displaystyle \rm = \lim_{ \theta \to \frac{\pi}{2} } \dfrac{ - 3\sin^{2} \theta \cos \theta }{ - 2 \sin \theta \cos\theta }$

$\displaystyle \rm = \lim_{ \theta \to \frac{\pi}{2} } \dfrac{3}{2} \cdot \sin \theta$

$\displaystyle \rm = \dfrac{3}{2} \cdot \sin \frac{\pi}{2}$

$\displaystyle \rm = \dfrac{3}{2} \cdot 1$

$\displaystyle \rm = \dfrac{3}{2}$

Therefore:

$\displaystyle \rm \longrightarrow\lim_{ \theta \to \frac{\pi}{2} } \bigg( \dfrac{1 - \sin^{3} \theta }{ { \cos}^{2} \theta } \bigg) = \dfrac{3}{2}$

Which is our required answer.

LEARN MORE:

$\displaystyle\rm 1.\:\: \lim_{x\to0}\sin(x)=0$

$\displaystyle\rm 2.\:\: \lim_{x\to0}\cos(x)=1$

$\displaystyle\rm 3.\:\: \lim_{x\to0}\dfrac{\sin(x)}{x}=1$

$\displaystyle\rm 4.\:\: \lim_{x\to0}\dfrac{\tan(x)}{x}=1$

$\displaystyle\rm 5.\:\: \lim_{x\to0}\dfrac{1-\cos(x)}{x}=0$

$\displaystyle\rm 6.\:\: \lim_{x\to0}\dfrac{\sin^{-1}(x)}{x}=1$

$\displaystyle\rm 7.\:\: \lim_{x\to0}\dfrac{\tan^{-1}(x)}{x}=1$

$\displaystyle\rm 8.\:\: \lim_{x\to0}\dfrac{\log(1+x)}{x}=1$

$\displaystyle\rm 9.\:\: \lim_{x\to0}\dfrac{e^{x}-1}{x}=1$