# Verify a limit using L'Hopital's Rule

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

This is what I did:

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

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

Delta2

## Answers and Replies

PeroK
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Not bad! Is there anything you're not sure about?

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

We can do that because $x^\left[(\ln a)/(1+ \ln x)\right]$ is continuous for x > 0, right?
.

Delta2
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Not quite, you can do that because the function ##\ln## is continuous.

murshid_islam
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.

DaveE, etotheipi, PeroK and 1 other person
PeroK
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I hope this part is correct:
$$\ln \left( \lim_{x \rightarrow 0^+} \left[x^\left[(\ln a)/(1+ \ln x)\right] \right] \right) = \lim_{x \rightarrow 0^+} \ln x^\left[(\ln a)/(1+ \ln x)\right]$$

We can do that because $x^\left[(\ln a)/(1+ \ln x)\right]$ 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!

etotheipi and Delta2
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}##

PeroK
Homework Statement:: Verify that $$\lim_{x \rightarrow 0^+} \left[x^\left[(\ln a)/(1+ \ln x)\right] \right]= a$$
Relevant Equations:: $$\lim_{x \rightarrow 0^+} \left[x^\left[(\ln a)/(1+ \ln x)\right] \right]= a$$

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

This is what I did:

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

Now, using L'Hopital's Rules, $$\ln y = (\ln a)\lim_{x \rightarrow 0^+} \frac{\frac{1}{x}}{\frac{1}{x}} = \ln a$$
Therefore, y = a
Was it necessary to show that the limit exists before letting $y = \lim_{x \rightarrow 0^+} \left[x^\left[(\ln a)/(1+ \ln x)\right] \right]$? Wasn't it going to be revealed at the end of the calculation if the limit didn't exist?

PeroK
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Was it necessary to show that the limit exists before letting $y = \lim_{x \rightarrow 0^+} \left[x^\left[(\ln a)/(1+ \ln x)\right] \right]$? 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.

If was taking the log of that equation that was the dubious step.
Why was it dubious?

PeroK
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Why was it dubious?
Because you don't know yet whether ##y## is a real number. See the example @Office_Shredder gave of why this can go wrong.

See the example @Office_Shredder gave of why this can go wrong.
But in that example, g(x) was not continuous.

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.

Delta2 and murshid_islam
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)?

PeroK
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: $$\lim \ln (s_n) = \ln a \ \Rightarrow \ \lim \exp(\ln (s_n)) = \exp (\ln a) \ \Rightarrow \ \lim s_n = a$$