What is the limit of this function as x approaches infinity?

Question

dumbsearch2 · Answer

Is there any picture you forgot to post?

anonymous · Answer

attachment

anonymous · Answer

no i was adding attachment

experimentx · Answer

method #1 multiply by conjuage
method #2 (take x^2) and expand the second term binomailly (or taylor expansion) and do some algebra, you should get answer.

experimentx · Answer

*conjugate

anonymous · Answer

I'm stuck at the algebra part here $$8x \over \sqrt{x^2+8x}-x$$. I know I have to devide by the highest denominator but what would that be x^2?

experimentx · Answer

whoops!! sorry ... 
change x-> -x and take limit x->\infty instead of -infty and try to use same method again.

anonymous · Answer

yeah i multiplied (radical x^2 +8x)-x and that's what i get ^

experimentx · Answer

no i meant let y=-x
and limit y->infty

ganeshie8 · Answer

divide x may be

experimentx · Answer

use the same trick, you will not have x-sqrt(x^2 ... ) down there, instead you will get +

anonymous · Answer

@ganeshie8 i'm not sure how to do this could you should me the result for (radical x^2+8x)/x @experimentX you mean in the beginning i should invert it?

experimentx · Answer

yes the limit is equivalent to 
$$ \lim_{y\to \infty } \sqrt{y^2 - 8y} - y$$

ganeshie8 · Answer

^we will do exactly as experimentX says hez the god of math

experimentx · Answer

lol ... who told ya that?

ganeshie8 · Answer

i sense that myself , seriously

experimentx · Answer

i'm pretty dumb for my own age, i sense that you guys are still in high school or uni freshman.

anonymous · Answer

so $$-8y \over \sqrt{y^2-8y} - y$$

experimentx · Answer

no ...

experimentx · Answer

you made some mistake

ganeshie8 · Answer

$\large \lim_{y -> \infty} \frac{-8y}{\sqrt{y^2 - 8y} + y} $

anonymous · Answer

oh oops +x sorry

experimentx · Answer

well well .. that solves it.

anonymous · Answer

but they say the answer is -4 and i'm not sure how to get that lol

ganeshie8 · Answer

we can divide y now top and bottom

experimentx · Answer

do it properly ... you will get your answer

anonymous · Answer

I understand that but the difficulty I am having is what to devide them by to get the limit

experimentx · Answer

y/y=1
sqrt(y^2 - 8y)/y = 1 (if you take the limit y-> \infty)

ganeshie8 · Answer

elaborating it


 $\large \lim_{y -> \infty} \frac{-8y}{\sqrt{y^2 - 8y} + y} $

 $\large \lim_{y -> \infty} \frac{-8}{\sqrt{1 - \frac{8}{y}} + 1} $

 $\large \lim_{\frac{1}{y} -> 0} \frac{-8}{\sqrt{1 - \frac{8}{y}} + 1} $

experimentx · Answer

yep that it is.

anonymous · Answer

yes got it thank you :)