Question

evaluate the limit using l hospitals rule if needed

Answer 1

\[\lim_{x \rightarrow \infty} \frac{2x-6}{4x^2+8x+5} \]

Answer 2

just look at the degrees of the numerator and denominator... as x approaches infinity, which gets bigger faster, the numerator or denominator?

Answer 3

denom

Answer 4

so it's like a small number divided by a very big number.. what's the value gonna be close to?

Answer 5

0

Answer 6

yes, zero is your limit...
but i bet your gonna say how do you do the algebra huh?

Answer 7

yessir :P

Answer 8

could we do like in the last question and ddivide all by x^2?

Answer 9

yes... multiply both the top and bottom by 1/x^2... and btw, i don't think ive done this with u...

Answer 10

sorry I thought you saw my last q :P

Answer 11

is there another green owl ur mistaking me with? there's an imposter!

Answer 12

BAN!

Answer 13

yep!

Answer 14

so now we have
\[\frac{\frac{2}{x}-\frac{6}{x^2}}{4+\frac{8}{x}+\frac{5}{x^2}}\]

Answer 15

correct... now just take the limit using the fact that
\(\huge \lim_{x\rightarrow \infty}\frac{c}{x^p}=0 \)

Answer 16

we hacve thathere x^p? :p

Answer 17

p>0

Answer 18

p and c are just constants...

Answer 19

yeah...

Answer 20

well I guess it would come out to 0/0 huh?

Answer 21

which I thought cannot be there...

Answer 22

either*

Answer 23

no, look at the lone 4...

Answer 24

yeah I saw that too
then 0/4?

Answer 25

you'll have \(\large \frac{0-0}{4+0+0} =0/4=0\)

Answer 26

okies ty :D

Answer 27

yw...:)