Verify a limit using L'Hopital's Rule

[murshid_islam]

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

 

[PeroK]

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

 

[murshid_islam]

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]

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

 

[Office_Shredder]

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.

 

[PeroK]

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!

 

[Delta2]

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 $f(x)=x^\frac{\ln a}{1+\ln x}$

 

[murshid_islam]

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]

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.

 

[murshid_islam]

If was taking the log of that equation that was the dubious step.

Why was it dubious?

 

[PeroK]

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.

 

[murshid_islam]

See the example @Office_Shredder gave of why this can go wrong.

But in that example, g(x) was not continuous.

 

[PeroK]

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.

 

[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]

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!