Limits

May 2010
15
0
Need helps solving these. I have no clue how to do them. (Thinking)








 

matheagle

MHF Hall of Honor
Feb 2009
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The first one is a zero times an infinity, so put the sine x in the denominator and make it infinity over infinity.

Actually I would just use sin x over x goes to 1...

\(\displaystyle \left({\sin x\over x}\right)\left({\ln 4+\ln x\over x^{-1}}\right)\sim -x\to 0\)

The second one uses conjugates.

\(\displaystyle \sqrt{x^2+3x}-x = {(\sqrt{x^2+3x}-x)(\sqrt{x^2+3x}+x)\over \sqrt{x^2+3x}+x}\)

\(\displaystyle = {3x\over \sqrt{x^2+3x}+x}\)

Now divide by x and take your limit, but be careful when passing the x though the square root.
3/2



The third one is \(\displaystyle e^{-4}\) by a theorem.
But it's easy to prove by taking the logarithm and using Lopies' Rule
 
Last edited:

CaptainBlack

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Nov 2005
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Need helps solving these. I have no clue how to do them. (Thinking)

rewrite:

\(\displaystyle \lim_{x\to 0^+} \sin(x) \ln(4x)=\lim_{x\to 0^+} \frac{\ln(4x)}{[1/\sin(x)]}\)

and apply L'Hopitals's rule.

CB
 

matheagle

MHF Hall of Honor
Feb 2009
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I used sin x over x immediately.