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Limits w/ l'hospital

  • Thread starter endeavor
  • Start date
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1. Homework Statement
Compute the limit [tex]\lim_{x \rightarrow 0} (1 + 3x)^{3/x}[/tex].

2. Homework Equations
3. The Attempt at a Solution

[tex]\lim_{x \rightarrow 0} (1 + 3x)^{3/x} = \lim_{x \rightarrow 0} e^{\frac{3}{x} \ln (1 + 3x)}[/tex]
[tex]\lim_{x \rightarrow 0} \frac{3}{x} \ln (1 + 3x) = 3 \lim_{x \rightarrow 0} \frac{ln (1 + 3x)}{x}[/tex]
[tex]= 3 \lim_{x \rightarrow 0} \frac{\frac{3}{1 + 3x}}{1}[/tex]
[tex]= 3 \lim_{x \rightarrow 0} \frac{3}{1 + 3x}[/tex]
[tex]=9[/tex]
[tex]\lim_{x \rightarrow 0} (1 + 3x)^{3/x} = e^9[/tex]

I just need to know if this is right.
 
Last edited:

Answers and Replies

Dick
Science Advisor
Homework Helper
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It is just fine.
 
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The answer is right. Just plug in .01 or .001 into the limit using the calculator and compare it to the exact value.
 
Gib Z
Homework Helper
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Or You could have manipulated the definition of the exponential: [tex]e^a=\lim_{x \rightarrow \inf} (1+\frac{a}{x})^x[/tex]

Rewrite [tex]\lim_{x \rightarrow 0} (1 + 3x)^{3/x}[/tex] as [tex]\lim_{x \rightarrow 0} (1 + \frac{3}{1/x})^{3(1/x)}[/tex]
The limit of 1/x as 0 approaches zero is the same as x appracohes infinity. So we can rewrite the limit as
[tex]\lim_{x \rightarrow \inf} (1 + 3/x)^{3x} = \lim_{x \rightarrow \inf}( (1 + 3/x)^x )^3[/tex] which is the definiton of e^3, then cubed, getting out e^9 as required.
 
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