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Finding the limit as $x$ approaches to infinity using L'Hopitals rule

$${\lim_{x \to \infty}{8x^{\frac{\ln2}{1+\ln x}}=L}}$$

$${\implies \lim_{x \to \infty}\frac{\ln 2}{1+\ln x}\ln8x=\ln L}$$

$$ = \ln2\lim_{x \to \infty}\frac{\ln8x}{1+\ln x} =\ln L$$

Now, I apply L'Hopitals rule

$${\implies \ln2\lim_{x \to \infty}}1=\ln L$$

This implies that ${L=2}$, however, this is not the case.

What is wrong?

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  • $\begingroup$ What's the derivative of $\ln 8x$? Oh, oops, mistake on my part. $\endgroup$
    – GFauxPas
    Nov 2 '14 at 22:01
  • $\begingroup$ It should be $\frac{8}{8x}=\frac{1}{x}$ $\endgroup$
    – Jason
    Nov 2 '14 at 22:02
  • $\begingroup$ The last implication does not contain an error. $\endgroup$
    – vadim123
    Nov 2 '14 at 22:02
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The error is in the very first implication. $$8x^a\neq (8x)^a$$ hence $$\ln 8x^a=\ln 8 +\ln x^a=\ln 8+a\ln x$$

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The mistake probably lies in your notation because the limit you provided is ambiguous. If you intended: $${\lim_{x \to \infty}{(8x)^{\frac{\ln2}{1+\ln x}}=L}}$$ then the limit is indeed 2. However, if you intended: $${\lim_{x \to \infty}{8(x^{\frac{\ln2}{1+\ln x}})=L}}$$ then you should have factored the 8 out, evaluated the rest of the limit using the steps you provided above, got 2, and obtained the final answer of 16 after multiplying by 8.

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$$x^{\frac{\log2}{1+\log x}}=\exp\left(\frac{\log2}{1+\log x}\log x\right)\;,\;\;\text{and}$$

$$\lim_{x\to\infty}\frac{\log x}{1+\log x}=1\stackrel{\text{continuity}}\implies x^{\frac{\log2}{1+\log x}} \xrightarrow[x\to\infty]{}e^{\log2}=2$$

and thus $\;L=16\;$

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