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Let $f_n\left(x\right)=n\left(\sqrt[n]{x}-1\right)$. How can I show that this function converges pointwise to $f\left(x\right)=\ln x$ on $\left(0,\infty\right)$? This is what I tried to do: $$\lim_{n\to\infty}f_n\left(x\right)=\lim_{n\to\infty}n\left(\sqrt[n]{x}-1\right)=\lim_{n\to\infty}n\sqrt[n]{x}-\lim_{n\to\infty}n.$$ I know that both terms approach infinity because the radical approaches $1$. However, I have no clue how to make the jump from that last expression to a logarithm. Is this a definition problem, or can I reach that conclusion algebraically? Thanks!

Edit 1: I have been fooling around with the equalities $\ln e^x=x$ and $e^{\ln x}=x$. I am still stuck, nevertheless. This is what I tried: $$n\left(\sqrt[n]{x}-1\right)=e^{\ln n\left(\sqrt[n]{x}-1\right)}=e^{\ln n+\ln\left(\sqrt[n]{x}-1\right)}.$$I cannot seem to progress further.

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Hint: L'Hopital. –  David Mitra Mar 12 '12 at 2:53
    
Well, have you tried to use $$e^x=\lim(1+x/n)^n$$? –  checkmath Mar 12 '12 at 3:10
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you might also be able to show $f_n'(x) \rightarrow 1/x$ and $f_n(1)=\ln 1=0$ (to avoid L'Hopital and more clearly see what is happening). –  ShawnD Mar 12 '12 at 3:30

2 Answers 2

up vote 1 down vote accepted

You could use L'Hôpital's rule. For $x\ne1$: $$ \lim_{n\rightarrow\infty} \bigl[ n (\root n\of x-1)\bigr]= \lim_{n\rightarrow\infty} {x^{1/n}-1\over 1/n}= \lim_{n\rightarrow\infty} {\ln x \cdot{-1\over n^2} \cdot x^{1/n} \over -1/n^2}= \lim_{n\rightarrow\infty} \bigl[{\ln x\cdot x^{1/n} }\bigr]=\ln x $$

Of course, for $x=1$, the limit is $0=\ln 1$.

Alternatively, you could write $$ \lim_{n\rightarrow\infty} {x^{1/n}-1\over 1/n} =\lim_{h\rightarrow 0^+} {x^{h}-1\over h}$$ Then the limit is recognized as being equal to the derivative of $x^t$ with respect to $t$ evaluated at $t=0$.

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Thank you! But I did not understand one thing: Why do we have to separate it into two cases, where $x\neq1$ and $x=1$? EDIT: Never mind, I saw it: It is because the limit is $0$. –  Josué Mar 12 '12 at 3:23
    
@JosuéMolina I suppose the argument would go through in the $x=1$ case. It seems an ugly way to do it though. –  David Mitra Mar 12 '12 at 3:32

If $x=1$, then $f_n(1)=0=\ln 1$ for each $n\in \mathbb{N}$.

Now assume $x\neq 1$. Since $x>0$ you have $x=e^{\ln x}$ and $\ln x\neq 0$, thus: $$\lim_{n\to \infty} n\ (x^{1/n}-1) = \lim_{n\to \infty} \frac{e^{(\ln x)/n}-1}{1/n} =\lim_{n\to \infty} \ln x\ \frac{e^{(\ln x)/n}-1}{(\ln x)/n}=\ln x\; ,$$ because of the fundamental limit $\displaystyle \lim_{y\to 0} \frac{e^y -1}{y} =1$.

Therefore $\displaystyle \lim_{n\to \infty} f_n(x)=\ln x$ for $x\in ]0,\infty[$.

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