anonymous
  • anonymous
limits of infinity
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
|dw:1371422236184:dw|
anonymous
  • anonymous
need your help ..people..
anonymous
  • anonymous
@julian25

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anonymous
  • anonymous
the denominator grows so much compared to the numerator when x -> INF
anonymous
  • anonymous
u can see that the grade of the denominator is 1 more than the numerator so the limit is 0
Jhannybean
  • Jhannybean
You'll need to use L''Hospital's rule here, I think since you get inf/inf
anonymous
  • anonymous
how did it happen?? can you please show me...??..pls...pls..
Jhannybean
  • Jhannybean
\[\large \lim_{x \rightarrow \infty }\frac{4x-5}{x^2-1}= \frac{4(\infty)-5}{(\infty)^2-1}= \frac{\infty}{\infty}\]
anonymous
  • anonymous
divide everything by x^2
Jhannybean
  • Jhannybean
You're not really supposed to plug them in but analyze their behavior lol.
anonymous
  • anonymous
$$\lim_{x\to\infty}\frac{x^2+1}{x^2}=1$$and $$\lim_{x\to\infty}\frac{4x-5}{4x}=1$$so we see \(x^2+1,x^2\) and \(4x-5,4x\) are asymptotically equivalent. We can then allow:$$\lim_{x\to\infty}\frac{4x}{x^2}=\lim_{x\to\infty}\frac4x=0$$
Mimi_x3
  • Mimi_x3
\[\lim_{x \rightarrow \inf} \frac{4x-5}{x^2+1}\] \[\lim_{x \rightarrow \inf}\frac{4x/x^2 -5/x^2}{x^2/x^2 +1/x^2} => \lim_{x \rightarrow \inf} \frac{4/x - 5/x^2}{1+ 1/x^2} =\frac{0}{1} =0\]
anonymous
  • anonymous
@Jhannybean asymptotic analysis is very important
anonymous
  • anonymous
thank you guys for your help ........... now I know how.......

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