Question

Evaluate the following limits. If needed, enter 'infinity' for \infty and '-infinity' for -\infty.
\lim_{ x \rightarrow \infty } \frac {\sqrt{ 10 + 9 x^2 } }{ 5 + 7 x } =
\lim_{ x \rightarrow -\infty } \frac {\sqrt{ 10 + 9 x^2 } }{ 5 + 7 x } =

Answer 1

\[\lim_{ x \rightarrow \infty } \frac {\sqrt{ 10 + 9 x^2 } }{ 5 + 7 x } =\]

Answer 2

isnt it the first one is infinity? but the answer is wrong

Answer 3

\[\frac{\sqrt{9x^2}}{7x}=\frac{3x}{7x}=\frac{3}{7}\] is the eyeball way

Answer 4

Oh!!! just take the greatest power?!

Answer 5

yeah the other stuff is not important if you have a ratio and \(x\to \infty\)
you can do it a longer more involved way to make a math teacher happy, but there is no need for it

Answer 6

\[\lim_{ x \rightarrow -\infty } \frac {\sqrt{ 10 + 9 x^2 } }{ 5 + 7 x } =
\] you have to be a bit more careful, because \(x\to -\infty\) and so \(7x<0\) whereas \(\sqrt{9x^2}>0\) so the answer is not \(\frac{3}{7}\) but rather \(-\frac{3}{7}\)

Answer 7

Oh yea so the answer for the second one is -3/7

Answer 8

Yes!! thx

Answer 9

yw

Answer 10

what about \lim_{ x \to \infty } \frac{ 3 }{ e^x + 8 } =

Answer 11

@satellite73 I know that if the limit go with -infty is will be 3/8
but how about the positive infty

Answer 12

\[\lim_{ x \to \infty } \frac{ 3 }{ e^x + 8 }\]

Answer 13

\(e^x\) goes to infinity lickety split, so that limit is zero

Answer 14

and you are right with the first one, since \(\lim_{x\to -\infty}e^x=0\)

Answer 15

Yea!!!!!! oh thx

Answer 16

yw

Answer 17

but im still confuse y limx→−∞e^x=0

Answer 18

b/c e^x is always 0?

Answer 19

think about \(e^{-100}=\frac{1}{e^{100}}\) it is an amazingly small number (i.e. close to zero) since \(e^100\) is really really large

Answer 20

oh yea it is close to 0
kk i got it REALLY thanks!!!