# indeterminate

• Nov 13th 2011, 01:43 PM
delgeezee
indeterminate
Use analytical methods to evaluate the following problem.

$\lim_{x\rightarrow 4} \frac{^5\sqrt{6x+8}-2}{\frac{1}{x}+\frac{1}{4}}$

the answer to the limit is $-\frac{6}{5}$

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 + $\frac{1}{4}$ disappear

$\frac{(6x+8)^{1/5}-2}{x^{-1}+\frac{1}{4}}$ = $\frac{\frac{1}{5}(6x+8)^{-4/5}*6}{-x^{-2}}$ = $\frac{\frac{6}{5}(6x+8)^{-4/5}}{-x^{-2}}$

I make powers have the same denominator
$-\frac{6}{5} \lim_{x\rightarrow 4}[ \frac{(6x+8)^{-4/5}}{x^{-\frac{10}{5}}} ]$

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.

$-\frac{6}{5} \lim_{x\rightarrow 4}[ \frac{(6x+8)^{-4/5}}{x^{-\frac{10}{5}}} ]^-5$ = $-\frac{6}{5} \lim_{x\rightarrow 4}[ \frac{(6x+8)^{4}}{x^{10}} ]$

I observe the limit which will equal 1 and multiply by -6/5

$-\frac{6}{5} \lim_{x\rightarrow 4}[ \frac{(6x+8)^{4}}{x^{10}} ]$ = $-\frac{6}{5} *1$

= $-\frac{6}{5}$
• Nov 13th 2011, 01:50 PM
Siron
Re: indeterminate
That looks fine!

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