anonymous
  • anonymous
Find the limit of (ln(x+1))/log2 x as x approaches to infinity.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\lim_{x\to \infty}\frac{\ln(x+1)}{\ln(2x)}\]?
primeralph
  • primeralph
|dw:1377573481672:dw|
anonymous
  • anonymous
primer is right. how do I do this?

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anonymous
  • anonymous
change 'o base formula
anonymous
  • anonymous
How?
primeralph
  • primeralph
Not PRIMER!!!!!
primeralph
  • primeralph
|dw:1377573631339:dw|
anonymous
  • anonymous
\[\log_2(x)=\frac{\ln(x)}{\ln(2)}\] right?
anonymous
  • anonymous
and it should be more or less obvious that \[\lim_{x\to \infty}\frac{\ln(x+1)}{\ln(x)}=1\]
primeralph
  • primeralph
@Idealist Got it?
anonymous
  • anonymous
Wait a minute.
anonymous
  • anonymous
But how did you get ln2 after ln2*ln(x+1)/lnx?
primeralph
  • primeralph
|dw:1377574035051:dw|
anonymous
  • anonymous
Thanks.
primeralph
  • primeralph
You're welcome.
anonymous
  • anonymous
I didnt get it prime... why is 1?
primeralph
  • primeralph
|dw:1377574273810:dw|
anonymous
  • anonymous
because the one is a red herring same reason \[\lim_{x\to \infty}\frac{x+1}{x}=1\]
anonymous
  • anonymous
the one doesn't contribute anything
primeralph
  • primeralph
Okay, no need for an elaborate proof then.
anonymous
  • anonymous
Ah... so u are using the limits properties of multiplication of limits right?

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