Question

Ok, sorry guys, but I have another calculus question:
How on earth do I do 11?
Why is number 9 on this worksheet negative root 2 over 2 and not positive root 2 over 2?
http://cdn.kutasoftware.com/Worksheets/Calc/01%20-%20Limits%20at%20Infinity.pdf

Answer 1

Ignore the thing about 9

Answer 2

I forgot i copied and pasted that, my question is about 11

Answer 3

For the lazy: evaluate\[\lim_{x\to-\infty}\frac{\sqrt{2x^2+3}}{2x+3}\]

Answer 4

By L'Hosiptal Rule, take the derivatives of the first term, u get \[-\frac{ 1/x }{ 4x^3 }=-\frac{ 1 }{ 4x^4 }\]
When x approaches to positive infinity, the term becomes 0.
And the second term,1, is a constant whatever x is. So the final answer is 1.

Answer 5

We didn't get to whatever that rule is yet

Answer 6

We are on chapter 2.1-2.2 in our book which is like the basics of limits and limits involving infinity

Answer 7

it is definitely not one

Answer 8

ooops which one are you doing?

Answer 9

if it is 11 then the answer is one as
\[\frac{\ln(x)}{x^4}\to 0\] lickety split

Answer 10

Sorry, it wasn't letting me respond from my phone. It is 11. I don't understand what you did

Answer 11

i guess i did nothing, maybe not good enough for a proof, but the log grows very very slowly, much much slower than \(x\) and certainly tons slower than \(x^4\)

Answer 12

so i just asserted that
\[\lim_{x\to \infty}\frac{\ln(x)}{x^4}=0\]

Answer 13

So you just have to have it memorized that ln x grows really slowly?

Answer 14

i would not call that "memorize" i would call it common sense
think about the common log (base 10)
to go from 3 to 4 you have to go from 1000 to 10000 a ten fold increase just to increase the output by a mere unit
or think about what
\[\frac{\log(x)}{x}\] is for \(x=1,000,000\)
it would be
\[\frac{6}{1,000,000}\] almost zero
and now imagine what it would be for \(\frac{\log(x)}{x^4}\) at \(x=1,000,000\)
up would have
\[\frac{6}{10^{12}}\] !!

Answer 15

Ok, so I forgot how logs worked