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DLS
 2 years ago
Limits
DLS
 2 years ago
Limits

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DLS
 2 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{x \rightarrow \infty }\frac{x^4+2x^3+3}{2x^4x+2}\]

DLS
 2 years ago
Best ResponseYou've already chosen the best response.0x+1 is a factor of this right?

TomLikesPhysics
 2 years ago
Best ResponseYou've already chosen the best response.1I 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.

TomLikesPhysics
 2 years ago
Best ResponseYou've already chosen the best response.1If you would use the L`Hopital Rule you would end up with it.

TomLikesPhysics
 2 years ago
Best ResponseYou've already chosen the best response.1Do you want to see the calculation?

DLS
 2 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{x \rightarrow \infty }\frac{3x^4+4x^3+}{4x^4+x^2+7}\] I'm trying to do this with the same concept but i'm getting 5/7 @TomLikesPhysics ?

DLS
 2 years ago
Best ResponseYou've already chosen the best response.0\[\frac{12x^3+12x^2}{16x^3+2x}\]

TomLikesPhysics
 2 years ago
Best ResponseYou've already chosen the best response.1the one where you got 5/7

DLS
 2 years ago
Best ResponseYou've already chosen the best response.0but how is it gaining infinity/infinity?

TomLikesPhysics
 2 years ago
Best ResponseYou've already chosen the best response.1How it gains infinity? You take the limit where x goes to infinity. So both terms grow. Is that what you meant?

DLS
 2 years ago
Best ResponseYou've already chosen the best response.0it is something+infinity is it samething?

DLS
 2 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{x \rightarrow \infty } \frac{x^3+x^2+4}{x^2+8}\] are you gettng 1 here?

TomLikesPhysics
 2 years ago
Best ResponseYou've already chosen the best response.1Here 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.

DLS
 2 years ago
Best ResponseYou've already chosen the best response.0No,I solved according to what you said and got this: \[\frac{3x^2+2x}{2x}\] so I got 3x+1=1?

TomLikesPhysics
 2 years ago
Best ResponseYou've already chosen the best response.1How do you end up with 3x+1=1 if you got a fraction?

DLS
 2 years ago
Best ResponseYou've already chosen the best response.0\[\frac{3x^2}{2x} + \frac{2x}{2x}\]

DLS
 2 years ago
Best ResponseYou've already chosen the best response.0I ended up with: \[\frac{3}{2}x+1\]

TomLikesPhysics
 2 years ago
Best ResponseYou've already chosen the best response.1k but if you got 3x+1 and let x approach infinity (taking the limit) than you end up with infinity.

DLS
 2 years ago
Best ResponseYou've already chosen the best response.0okay! if it was 1/x and x==> infinity it would have been 1 RIGHT?

TomLikesPhysics
 2 years ago
Best ResponseYou've already chosen the best response.1limit 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.

TomLikesPhysics
 2 years ago
Best ResponseYou've already chosen the best response.1If you got 1/x and x gets bigger and bigger you would get a limit which is 0.

TomLikesPhysics
 2 years ago
Best ResponseYou've already chosen the best response.11/x as x approaches infinity is 0. 1/x as x approaches 0 is infinity.

DLS
 2 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{x \rightarrow 0} \frac{\cos7x\cos9x}{\cos3x\cos5x}\] These ones?

TomLikesPhysics
 2 years ago
Best ResponseYou've already chosen the best response.1I don´t know. Have not seen this one before.^^ Never dealt with limits and sine or cosine.

DLS
 2 years ago
Best ResponseYou've already chosen the best response.0okay..book did a direct step after this : \[\frac{2\sin8xsinx}{2\sin4xsinx} \]

TomLikesPhysics
 2 years ago
Best ResponseYou've already chosen the best response.1^^ First time I see that kind of stuff. I´m sorry but with that I can not help you.
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