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I tried:

$$\lim_{x \rightarrow 0} \frac{\ln (x+1)}{\ln (2x+1)} = \frac{1}{\ln(x+1)} \cdot \ln (2x+1) = \ln (2x+1)^{\frac{1}{\ln(x+1)}} = ???$$

What do I do next? I feel like I could use $\lim_{x \rightarrow 0}\frac{\ln (x+1)}{x}=1$ but I am not sure how.

1 Answers 1

6

Hint:

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

  • 0
    What did you do? – 2017-02-22
  • 0
    ...ln is not sin; you might want to tweak those terms... – 2017-02-22
  • 0
    @MarkRead Note that the fractions on the right all cancel out to give you the left side. – 2017-02-22
  • 2
    @SimplyBeautifulArt While what you've written isn't wrong, the second multiplicand on the right doesn't have a finite limit when $x \to 0$. – 2017-02-22
  • 1
    @B.Mehta Ah, right. XD – 2017-02-22
  • 0
    Could you help me further? I got $\frac{\ln(x+1)}x\frac{x}{\ln(2x+1)}$, what do I do next? If I cancel the $x$ out I get back to square one – 2017-02-22
  • 0
    Actually I still need help. $\frac{\ln(x+1)}x\frac{x}{\ln(2x+1)}$ tends to zero and the right answer is .5. What did I do wrong? – 2017-02-22
  • 0
    Nevermind, I got it. I'd like to know how you knew that you had to multiply by $\frac{2x}{2x}\cdot \frac{x}{x}$ though. – 2017-02-22
  • 0
    @MarkRead Practice. We know that $\frac{\ln(1+x)}x\to1$, and so we try to use this to our fullest – 2017-02-22
  • 1
    Also, good job solving it on your own (while I was a tad busy, sorry for late reply) – 2017-02-22
  • 0
    Right, thanks again. – 2017-02-22