DLS
  • DLS
Limits
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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DLS
  • DLS
\[\lim_{x \rightarrow \infty }\frac{x^4+2x^3+3}{2x^4-x+2}\]
DLS
  • DLS
x+1 is a factor of this right?
DLS
  • DLS
however,no use..

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More answers

anonymous
  • anonymous
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.
DLS
  • DLS
it is 1/2 only but how
anonymous
  • anonymous
If you would use the L`Hopital Rule you would end up with it.
anonymous
  • anonymous
Do you want to see the calculation?
DLS
  • DLS
\[\lim_{x \rightarrow \infty }\frac{3x^4+4x^3+}{4x^4+x^2+7}\] I'm trying to do this with the same concept but i'm getting 5/7 @TomLikesPhysics ?
DLS
  • DLS
+5
DLS
  • DLS
*
DLS
  • DLS
\[\frac{12x^3+12x^2}{16x^3+2x}\]
anonymous
  • anonymous
should be 3/4
DLS
  • DLS
??
anonymous
  • anonymous
the one where you got 5/7
DLS
  • DLS
okay got it!
DLS
  • DLS
but how is it gaining infinity/infinity?
anonymous
  • anonymous
How it gains infinity? You take the limit where x goes to infinity. So both terms grow. Is that what you meant?
DLS
  • DLS
it is something+infinity is it samething?
DLS
  • DLS
\[\lim_{x \rightarrow \infty } \frac{x^3+x^2+4}{x^2+8}\] are you gettng 1 here?
anonymous
  • anonymous
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.
DLS
  • DLS
No,I solved according to what you said and got this: \[\frac{3x^2+2x}{2x}\] so I got 3x+1=1?
anonymous
  • anonymous
How do you end up with 3x+1=1 if you got a fraction?
DLS
  • DLS
\[\frac{3x^2}{2x} + \frac{2x}{2x}\]
DLS
  • DLS
I ended up with: \[\frac{3}{2}x+1\]
anonymous
  • anonymous
k but if you got 3x+1 and let x approach infinity (taking the limit) than you end up with infinity.
DLS
  • DLS
okay! if it was 1/x and x==> infinity it would have been 1 RIGHT?
anonymous
  • anonymous
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.
DLS
  • DLS
Great.Clear now :)
anonymous
  • anonymous
If you got 1/x and x gets bigger and bigger you would get a limit which is 0.
DLS
  • DLS
yeah!
anonymous
  • anonymous
1/x as x approaches infinity is 0. 1/x as x approaches 0 is infinity.
DLS
  • DLS
\[\lim_{x \rightarrow 0} \frac{\cos7x-\cos9x}{\cos3x-\cos5x}\] These ones?
anonymous
  • anonymous
I don´t know. Have not seen this one before.^^ Never dealt with limits and sine or cosine.
DLS
  • DLS
okay..book did a direct step after this :| \[\frac{2\sin8xsinx}{2\sin4xsinx} \]
anonymous
  • anonymous
^^ First time I see that kind of stuff. I´m sorry but with that I can not help you.

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