
indeterminate
Use analytical methods to evaluate the following problem.
$\displaystyle \lim_{x\rightarrow 4} \frac{^5\sqrt{6x+8}2}{\frac{1}{x}+\frac{1}{4}}$
the answer to the limit is $\displaystyle \frac{6}{5}$
I have been taking a stab at it for quite a while. I think I figured it out but I a want to make sure the methods to how I got the answer were proper.
First thing, Apply L' Hoptial's rule to make the 2 and the +$\displaystyle \frac{1}{4}$ disappear
$\displaystyle \frac{(6x+8)^{1/5}2}{x^{1}+\frac{1}{4}}$ = $\displaystyle \frac{\frac{1}{5}(6x+8)^{4/5}*6}{x^{2}}$ = $\displaystyle \frac{\frac{6}{5}(6x+8)^{4/5}}{x^{2}} $
I make powers have the same denominator
$\displaystyle \frac{6}{5} \lim_{x\rightarrow 4}[ \frac{(6x+8)^{4/5}}{x^{\frac{10}{5}}} ]$
I get rid of the exponent denominator by applying an exponent of the power 5 to the limit. I also make it positive for simplicity.
$\displaystyle \frac{6}{5} \lim_{x\rightarrow 4}[ \frac{(6x+8)^{4/5}}{x^{\frac{10}{5}}} ]^5$ = $\displaystyle \frac{6}{5} \lim_{x\rightarrow 4}[ \frac{(6x+8)^{4}}{x^{10}} ]$
I observe the limit which will equal 1 and multiply by 6/5
$\displaystyle \frac{6}{5} \lim_{x\rightarrow 4}[ \frac{(6x+8)^{4}}{x^{10}} ]$ = $\displaystyle \frac{6}{5} *1$
=$\displaystyle \frac{6}{5} $

Re: indeterminate
That looks fine!
Notice the denominator of the limit it has to be $\displaystyle \frac{1}{x}\frac{1}{4}$ else you don't have the indeterminate form $\displaystyle \frac{0}{0}$ but offcourse it doesn't make a difference to apply l'Hopital's rule.