Limit with logarithms (no l'Hospital)

Question

I have following limit: $$\lim_{x \to +\infty} \ (3x+1)(\log(2x)-\log(2x-1))$$ I've tried to do this: $$\lim_{x \to +\infty} \ (3x+1)(\log(2x)-\log(2x-1))=\lim_{x \to +\infty} \ \log(\frac{2x}{2x-1})^{(3x+1)}=\lim_{x \to +\infty} \ \log(1-2x)^{(3x+1)}$$ I think I should use this equation next: $$\lim_{x \to 0} \ (1+x)^{\frac{1}{x}}=e$$ But I don't know how it it will interact with $\log$. Can anybody help me with that?

Answer 1

Let us denote the above limit by $L$. Then we have that $$\begin{eqnarray} -L &=& \lim_{x\to\infty} (3x+1) \cdot \left( \log(2x-1) - \log(2x) \right) \\&=& \lim_{x\to\infty} (3x+1) \cdot \log\left( 1 - \frac{1}{2x} \right) \end{eqnarray}$$

Now we can use the Taylor expansion $$\log\left(1+\frac{1}{t}\right) = \frac{1}{t} + O\left(\frac{1}{t^2}\right).$$

Then

$$\begin{eqnarray} -L &=&\lim_{x\to\infty} (3x+1) \cdot \left( -\frac{1}{2x} + O(1/x^2) \right) \\&=& \lim_{x\to\infty} - \frac{3}{2} - \frac{1}{2x} + O(1/x) \\&=& -\frac{3}{2} \end{eqnarray}$$

or $L=\frac{3}{2}$.