Help!

The lack of image focus is giving me a headache. Also sideways

File: tmp_19808-20161129_004511-1844000338.jpg (3MB, 4032x3024px) Image search: [Google]

3MB, 4032x3024px

Shit. Sorry. Is this better?

File: tmp_19808-20161129_004627598607175.jpg (3MB, 4032x3024px) Image search: [Google]

3MB, 4032x3024px

Also there's a back. That's that side I did the worst on.

>>8503827
Better but still sideways.
I'm a little rusty but when you consider the limit of f(x) going to infinity, you need to consider the 'large terms' "only". In the case of problem 1, you only need to consider the x^2 terms in the numerator and denominator.

in part a of number 4 you didn't square one of the terms that you plug in for x=10.

Also wow at your teacher for labeling the problems as the square roots of the squares.

>>8503820
>has stupid question
>doesn't post it in SQT
>instead posts multiple large image files with shit focus on their side
No thanks.

Just think,
what would jesus do?

>>8503820
>>8503827
>>8503829
Should have paid attention in class faggot.

Not sure how you got your answer for part b of number 4 but you have to convert the 5 hours to minutes (x=300mins). That's a pretty common mistake students do

>>8503820
>>8503827
>>8503829
>period

wew lad

>>8503827
When you take the limit of something for x going to infinity, you should just divide every single term by the largest power of x. Like for question 2, you have [math] \frac{-7+2x}{-6x+x^2} [/math] so you divide everything by x^2 so you end up with [math] \frac{ \frac{-7}{x^2} + \frac{2}{x} }{ \frac{-6}{x} + 1} [/math] If you plug in infinity now, you'll end up with (0+0)/(1+0) which is 0. This almost always works, and it's so easy to do in your head.

Just write a dick on your paper and turn it in