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Find limit of the function: $f(x) = \frac{\ln(x)}{\ln(2x)}$

Solve:

$\lim\frac{\ln(x)}{\ln(2x)} =\lim\frac{(\ln(x))'}{(\ln((2x))'}=\frac{\frac{1}{x}}{\frac{1}{2x}} = 2 $

However, the answer is $1$. What did I do wrong?

Winther
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Stoatman
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  • You have not specified $\lim_{x\to ?}$. I suppose it's $\infty$?! – Winther Jan 19 '16 at 16:34
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    Note that $\ln(2x) = \ln(x) + \ln(2)$ therefore $\ln(2x) = \ln(x) + C$ where $C=\ln(2)$. The functions $\ln(2x)$ and $\ln(x)$ have the same derivative, because they defer by a constant. – Joel Jan 19 '16 at 16:42

3 Answers3

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The derivative of $\ln(2x)$ is not $\frac{1}{2x}$ but $\frac{1}{x}.$

You might also explain what limit is in question. However, the answer will remain the same.

Marko
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Also, another way $$ \frac{\ln x}{\ln (2x)}= \frac{\ln x}{\ln x + \ln 2}=\frac{1}{1+\ln(2)/\ln(x)} $$ which tends to $1$ as $x\to\infty$, or as $x\to0$.

Brightsun
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You got the derivative of $\ln x$ wrong. Use the chain rule $u(v)' = v'u'(v)$:

$${d\over dx}\ln(2x) = 2\cdot{1 \over2x} = {1\over x}$$

adjan
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