Question

Limits

Answer 1

\[\lim_{x \rightarrow \infty }\frac{x^4+2x^3+3}{2x^4-x+2}\]

Answer 2

x+1 is a factor of this right?

Answer 3

however,no use..

Answer 4

I think it is one half.
Because you got up- and downstairs and x^4 which I think is a special case and in which the limit approaches the coefficents before these two identical powers of x which is 1upstairs and 2 downstairs - 1/2.

Answer 5

it is 1/2 only but how

Answer 6

If you would use the L`Hopital Rule you would end up with it.

Answer 7

Do you want to see the calculation?

Answer 8

\[\lim_{x \rightarrow \infty }\frac{3x^4+4x^3+}{4x^4+x^2+7}\]
I'm trying to do this with the same concept but i'm getting 5/7
@TomLikesPhysics ?

Answer 9

+5

Answer 10

*

Answer 11

\[\frac{12x^3+12x^2}{16x^3+2x}\]

Answer 12

should be 3/4

Answer 13

??

Answer 14

the one where you got 5/7

Answer 15

okay got it!

Answer 16

but how is it gaining infinity/infinity?

Answer 17

How it gains infinity?
You take the limit where x goes to infinity.
So both terms grow.
Is that what you meant?

Answer 18

it is something+infinity
is it samething?

Answer 19

\[\lim_{x \rightarrow \infty } \frac{x^3+x^2+4}{x^2+8}\]
are you gettng 1 here?

Answer 20

Here it should be infinity.
You got downstairs an x^2 and upstairs and x^3. So the x^3 "is stronger" and wins and therefore the whole thing goes to infinity.

Answer 21

No,I solved according to what you said and got this:
\[\frac{3x^2+2x}{2x}\]
so I got 3x+1=1?

Answer 22

How do you end up with 3x+1=1 if you got a fraction?

Answer 23

\[\frac{3x^2}{2x} + \frac{2x}{2x}\]

Answer 24

I ended up with:
\[\frac{3}{2}x+1\]

Answer 25

k
but if you got 3x+1 and let x approach infinity (taking the limit) than you end up with infinity.

Answer 26

okay!
if it was 1/x and x==> infinity it would have been 1 RIGHT?

Answer 27

limit x-> inf. of ax+b is alsways infinity. The b does not matter in the long run and nether does a while x gets bigger and bigger.

Answer 28

Great.Clear now :)

Answer 29

If you got 1/x and x gets bigger and bigger you would get a limit which is 0.

Answer 30

yeah!

Answer 31

1/x as x approaches infinity is 0.
1/x as x approaches 0 is infinity.

Answer 32

\[\lim_{x \rightarrow 0} \frac{\cos7x-\cos9x}{\cos3x-\cos5x}\]
These ones?

Answer 33

I don´t know. Have not seen this one before.^^
Never dealt with limits and sine or cosine.

Answer 34

okay..book did a direct step after this :|
\[\frac{2\sin8xsinx}{2\sin4xsinx} \]

Answer 35

^^
First time I see that kind of stuff. I´m sorry but with that I can not help you.