Question

find the limit. use lhospitals rule where appropriate. if there is a more elementary method, consider using it.
lim _ { \theta
ightarrow pi / 2 } \frac { 1 - sin ( \theta ) } { 1 + cos ( 2 \theta ) }
find the limit. use lhospitals rule if appropriate. if there is a more elementary method, consider using it.
lim _ { x
ightarrow infty } \frac { ln ( x ) } { sqrt { x } }

Explanation:

Step1: Check the form of the limit $\lim_{\theta\to\frac{\pi}{2}}\frac{1 - \sin(\theta)}{1+\cos(2\theta)}$

Substitute $\theta=\frac{\pi}{2}$ into the function:
When $\theta=\frac{\pi}{2}$, $1-\sin(\theta)=1 - \sin(\frac{\pi}{2})=1 - 1=0$
and $1+\cos(2\theta)=1+\cos(\pi)=1-1 = 0$. So, it is in the $\frac{0}{0}$ form.

Step2: Apply L'Hospital's Rule

Differentiate the numerator and the denominator.
The derivative of $y = 1-\sin(\theta)$ with respect to $\theta$ is $y^\prime=-\cos(\theta)$
The derivative of $y = 1+\cos(2\theta)$ with respect to $\theta$ is $y^\prime=- 2\sin(2\theta)$
So, $\lim_{\theta\to\frac{\pi}{2}}\frac{1 - \sin(\theta)}{1+\cos(2\theta)}=\lim_{\theta\to\frac{\pi}{2}}\frac{-\cos(\theta)}{-2\sin(2\theta)}$

Step3: Simplify and substitute $\theta=\frac{\pi}{2}$

Since $\sin(2\theta)=2\sin(\theta)\cos(\theta)$, then $\lim_{\theta\to\frac{\pi}{2}}\frac{-\cos(\theta)}{-2\sin(2\theta)}=\lim_{\theta\to\frac{\pi}{2}}\frac{-\cos(\theta)}{-4\sin(\theta)\cos(\theta)}$
Cancel out $\cos(\theta)$ (for $\theta
eq\frac{\pi}{2}+k\pi,k\in\mathbb{Z}$ near $\theta = \frac{\pi}{2}$), we get $\lim_{\theta\to\frac{\pi}{2}}\frac{1}{4\sin(\theta)}$
Substitute $\theta=\frac{\pi}{2}$, $\sin(\frac{\pi}{2}) = 1$

Step1: Check the form of the limit $\lim_{x\to\infty}\frac{\ln(x)}{\sqrt{x}}$

As $x\to\infty$, $\ln(x)\to\infty$ and $\sqrt{x}\to\infty$, so it is in the $\frac{\infty}{\infty}$ form.

Step2: Apply L'Hospital's Rule

Differentiate the numerator and the denominator.
The derivative of $y=\ln(x)$ with respect to $x$ is $y^\prime=\frac{1}{x}$
The derivative of $y = \sqrt{x}=x^{\frac{1}{2}}$ with respect to $x$ is $y^\prime=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}}$
So, $\lim_{x\to\infty}\frac{\ln(x)}{\sqrt{x}}=\lim_{x\to\infty}\frac{\frac{1}{x}}{\frac{1}{2\sqrt{x}}}$

Step3: Simplify and find the limit

$\lim_{x\to\infty}\frac{\frac{1}{x}}{\frac{1}{2\sqrt{x}}}=\lim_{x\to\infty}\frac{2}{\sqrt{x}}$
As $x\to\infty$, $\frac{2}{\sqrt{x}}\to0$

Answer:

$\frac{1}{4}$