I'm trying to evaluate the following limit:
$\displaystyle\lim_{x \to \infty} \displaystyle\frac{\ln(2x)}{\ln(x)}$
Using L'Hospital's rule, I end up with:
$\displaystyle\lim_{x \to \infty} \displaystyle\frac{\frac{1}{x}}{\frac{2}{x}}$
= $\displaystyle\lim_{x \to \infty} x/2x$
= $\displaystyle\lim_{x \to \infty} 1/2$
= $1/2$
Is this correct?
$\endgroup$ 53 Answers
$\begingroup$HINT: No differentiation is needed:
$$\frac{\ln 2x}{\ln x}=\frac{\ln 2+\ln x}{\ln x}=1+\frac{\ln 2}{\ln x}$$
You can use l’Hospital’s rule here, but you have to differentiate $\ln 2x$ correctly.
$\endgroup$ 1 $\begingroup$No. You forgot to apply the chain rule when differentiating $\ln 2x$.
Also, I'm not sure where $n$ came from.
$\endgroup$ $\begingroup$You answer is incorrect, and your should have checked this by experimentation. Putting $x=10^{10}$ gives $\ln(2x)/\ln x \approx 1.03$ while putting $x=10^{50}$ gives $\ln(2x)/\ln x \approx 1.01$. It would seem that $\ln(2x)/\ln x \to 1$ as $x \to \infty$. However, we need to prove this rigorously.
Simply apply the laws of logs: $\ln(ab) = \ln a + \ln b$. We see that
$$\frac{\ln (2x)}{\ln x} \equiv \frac{\ln 2 + \ln x}{\ln x} \equiv \frac{\ln 2}{\ln x}+1$$
Since $\ln x \to \infty$ as $x \to \infty$ it follows that $\ln(2x)/\ln x \to 1$ as $x \to \infty$.
If you insist upon using L'Hopital's rule then notice that:
\begin{array}{ccc} \frac{\operatorname{d}}{\operatorname{d}\!x}\ln(2x) &=& \frac{2}{2x} \equiv \frac{1}{x} \\ \frac{\operatorname{d}}{\operatorname{d}\!x}\ln(x) &=& \frac{1}{x} \end{array}
It follows that
$$\lim_{x \to \infty}\frac{\ln(2x)}{\ln x} = \lim_{x \to \infty}\frac{1/x}{1/x} = 1$$
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