Limitless 5

1. Carry out the division. $\displaystyle\lim_{x\rightarrow \inf} \frac{(2 \log_{4}3\sqrt{4x^7})}{\sqrt{x}} + \frac{\sin^2(\sqrt{x})}{\sqrt{x}}$

Apply squeeze theorem for $\displaystyle\frac{\sin^2(\sqrt{x})}{\sqrt{x}}$

$\sin(x)$ is ranging from -1 to 1, $\sin^2(x)$ is ranging from 0 to 1

$\displaystyle\lim_{x\rightarrow \inf} 0 < \displaystyle\lim_{x\rightarrow \inf} \sin^2(\sqrt{x}) < \displaystyle\lim_{x\rightarrow \inf}1$

Divide three sides by $\sqrt{x}$

$\displaystyle\lim_{x\rightarrow \inf} \frac{0}{\sqrt{x}} < \displaystyle\lim_{x\rightarrow \inf} \frac{\sin^2(\sqrt{x})}{\sqrt{x}} < \displaystyle\lim_{x\rightarrow \inf}\frac{1}{\sqrt{x}}$

$\displaystyle\lim_{x\rightarrow \inf}\frac{1}{\sqrt{x}}$ = 0 as well as $\displaystyle\lim_{x\rightarrow \inf} \frac{0}{\sqrt{x}} = 0$

$\therefore$ $\displaystyle\lim_{x\rightarrow \inf} \frac{\sin^2(\sqrt{x})}{\sqrt{x}}$ = 0

$\displaystyle\lim_{x\rightarrow \inf} \frac{(2 \log_{4}3\sqrt{4x^7})}{\sqrt{x}} + 0$

Use property of logarithm:

$\displaystyle\lim_{x\rightarrow \inf} \frac{ \log_{4}(3\sqrt{4x^7})^2}{\sqrt{x}} =$ $\displaystyle\lim_{x\rightarrow \inf} \frac{ \log_{4}(36x^7)}{\sqrt{x}}$

Use another property of logarithm:

$\displaystyle\lim_{x\rightarrow \inf} \frac{ \log_{4}(36) + \log_{4}x^7}{\sqrt{x}}$ = $\displaystyle\lim_{x\rightarrow \inf} \frac{ \log_{4}(36) + 7\log_{4}x}{\sqrt{x}}$

We have indeterminate form (infinity over infinity), therefore we can apply L'Hôpital's rule; however, there is another way.

Logarithmic functions grow slower than algebraic.

As $\sqrt{x}$ goes to infinity faster than $7\log_{4}(x)$ , $\displaystyle\lim_{x\rightarrow \inf} \frac{ \log_{4}(36) + 7\log_{4}x}{\sqrt{x}} = 0$

$\therefore \displaystyle\lim_{x\rightarrow \inf} \frac{(2 \log_{4}3\sqrt{4x^7})}{\sqrt{x}} + \frac{\sin^2(\sqrt{x})}{\sqrt{x}} = \boxed{0}$