Verify a limit using L'Hopital's Rule

  • #1
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Homework Statement:
Verify that [tex]\lim_{x \rightarrow 0^+} \left[x^\left[(\ln a)/(1+ \ln x)\right] \right]= a[/tex]
Relevant Equations:
[tex]\lim_{x \rightarrow 0^+} \left[x^\left[(\ln a)/(1+ \ln x)\right] \right]= a[/tex]
I have to prove that [itex]\lim_{x \rightarrow 0^+} \left[x^\left[(\ln a)/(1+ \ln x)\right] \right]= a[/itex] (in order to show that the indeterminate form of the type [itex]0^0[/itex] can be any positive real number).

This is what I did:

Let [tex]y = \lim_{x \rightarrow 0^+} \left[x^\left[(\ln a)/(1+ \ln x)\right] \right][/tex]
[tex]\ln y = \ln \left( \lim_{x \rightarrow 0^+} \left[x^\left[(\ln a)/(1+ \ln x)\right] \right] \right) = \lim_{x \rightarrow 0^+} \ln x^\left[(\ln a)/(1+ \ln x)\right] = \lim_{x \rightarrow 0^+} \frac{\ln a}{1+ \ln x} \ln x = (\ln a)\lim_{x \rightarrow 0^+} \frac{\ln x}{1+ \ln x}[/tex]

Now, using L'Hopital's Rules, [tex]\ln y = (\ln a)\lim_{x \rightarrow 0^+} \frac{\frac{1}{x}}{\frac{1}{x}} = \ln a[/tex]
Therefore, y = a
 

Answers and Replies

  • #2
PeroK
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Not bad! Is there anything you're not sure about?
 
  • #3
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Not bad! Is there anything you're not sure about?
I hope this part is correct:
[tex]\ln \left( \lim_{x \rightarrow 0^+} \left[x^\left[(\ln a)/(1+ \ln x)\right] \right] \right) = \lim_{x \rightarrow 0^+} \ln x^\left[(\ln a)/(1+ \ln x)\right][/tex]

We can do that because [itex]x^\left[(\ln a)/(1+ \ln x)\right][/itex] is continuous for x > 0, right?
.
 
  • #4
Delta2
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Not quite, you can do that because the function ##\ln## is continuous.
 
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  • #5
Office_Shredder
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One thing to watch out for is that's only true if the limit exists to begin with.

Here's a dumb counterexample. Let ##f(x)=x^2##, and let ##g(x)## be 1 if x is rational, -1 if x is irrational. Then
$$f(\lim_{x\to 1 }g(x))$$

Doesn't exist, but

$$\lim_{x\to 1} f(g(x))=1.$$

The fact that ln is monotonic is actually fairly important for concluding that your new limit exists implies the old limit exists also.
 
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  • #6
PeroK
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I hope this part is correct:
[tex]\ln \left( \lim_{x \rightarrow 0^+} \left[x^\left[(\ln a)/(1+ \ln x)\right] \right] \right) = \lim_{x \rightarrow 0^+} \ln x^\left[(\ln a)/(1+ \ln x)\right][/tex]

We can do that because [itex]x^\left[(\ln a)/(1+ \ln x)\right][/itex] is continuous for x > 0, right?
.
Also, you could work backwards. First you show that the limit on the right-hand side exists and equals ##\ln a##. You then use the fact that the exponential is a continuous function and take the exponential inside the limit. So, if ##s_n## is your sequence, you have: $$\lim \ln (s_n) = \ln a \ \Rightarrow \ \lim \exp(\ln (s_n)) = \exp (\ln a) \ \Rightarrow \ \lim s_n = a$$
PS By mistake I've used a sequence instead of a function, but the same technique applies!
 
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  • #7
Delta2
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Also, you could work backwards. First you show that the limit on the right-hand side exists and equals ##\ln a##. You then use the fact that the exponential is a continuous function and take the exponential inside the limit. So, if ##s_n## is your sequence, you have: $$\lim \ln (s_n) = \ln a \ \Rightarrow \ \lim \exp(\ln (s_n)) = \exp (\ln a) \ \Rightarrow \ \lim s_n = a$$
PS By mistake I've used a sequence instead of a function, but the same technique applies!
Very well, I also believe this is the absolutely correct way to do it. Your only glitch is that we don't have a sequence but a function ##f(x)=x^\frac{\ln a}{1+\ln x}##
 
  • #8
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Homework Statement:: Verify that [tex]\lim_{x \rightarrow 0^+} \left[x^\left[(\ln a)/(1+ \ln x)\right] \right]= a[/tex]
Relevant Equations:: [tex]\lim_{x \rightarrow 0^+} \left[x^\left[(\ln a)/(1+ \ln x)\right] \right]= a[/tex]

I have to prove that [itex]\lim_{x \rightarrow 0^+} \left[x^\left[(\ln a)/(1+ \ln x)\right] \right]= a[/itex] (in order to show that the indeterminate form of the type [itex]0^0[/itex] can be any positive real number).

This is what I did:

Let [tex]y = \lim_{x \rightarrow 0^+} \left[x^\left[(\ln a)/(1+ \ln x)\right] \right][/tex]
[tex]\ln y = \ln \left( \lim_{x \rightarrow 0^+} \left[x^\left[(\ln a)/(1+ \ln x)\right] \right] \right) = \lim_{x \rightarrow 0^+} \ln x^\left[(\ln a)/(1+ \ln x)\right] = \lim_{x \rightarrow 0^+} \frac{\ln a}{1+ \ln x} \ln x = (\ln a)\lim_{x \rightarrow 0^+} \frac{\ln x}{1+ \ln x}[/tex]

Now, using L'Hopital's Rules, [tex]\ln y = (\ln a)\lim_{x \rightarrow 0^+} \frac{\frac{1}{x}}{\frac{1}{x}} = \ln a[/tex]
Therefore, y = a
Was it necessary to show that the limit exists before letting [itex]y = \lim_{x \rightarrow 0^+} \left[x^\left[(\ln a)/(1+ \ln x)\right] \right][/itex]? Wasn't it going to be revealed at the end of the calculation if the limit didn't exist?
 
  • #9
PeroK
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Was it necessary to show that the limit exists before letting [itex]y = \lim_{x \rightarrow 0^+} \left[x^\left[(\ln a)/(1+ \ln x)\right] \right][/itex]? Wasn't it going to be revealed at the end of the calculation if the limit didn't exist?
If was taking the log of that equation that was the dubious step.
 
  • #10
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If was taking the log of that equation that was the dubious step.
Why was it dubious?
 
  • #13
PeroK
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But in that example, g(x) was not continuous.
Okay. Let's say we have: $$y = \lim_{x \rightarrow 0^+} x \ \ \text{and} \ \ \ln y = \lim_{x \rightarrow 0^+} \ln x$$
And you've effectively got an expression with the log of zero. This is why it's dubious, because ##\ln y## is actually undefined. Even if you end up concluding that ##y = 0##, the expressions you have are not mathematically robust.
 
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  • #14
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Okay. Let's say we have: $$y = \lim_{x \rightarrow 0^+} x \ \ \text{and} \ \ \ln y = \lim_{x \rightarrow 0^+} \ln x$$
And you've effectively got an expression with the log of zero. This is why it's dubious, because ##\ln y## is actually undefined. Even if you end up concluding that ##y = 0##, the expressions you have are not mathematically robust.
So how would you suggest I change my initial work (and do I need to)?
 
  • #15
PeroK
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So how would you suggest I change my initial work (and do I need to)?
Also, you could work backwards. First you show that the limit on the right-hand side exists and equals ##\ln a##. You then use the fact that the exponential is a continuous function and take the exponential inside the limit. So, if ##s_n## is your sequence, you have: $$\lim \ln (s_n) = \ln a \ \Rightarrow \ \lim \exp(\ln (s_n)) = \exp (\ln a) \ \Rightarrow \ \lim s_n = a$$
PS By mistake I've used a sequence instead of a function, but the same technique applies!
 
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