- #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

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