I have an issue solving this limit:

 

[latex]\lim_{x\to 0}\frac{\; ln\left (1+3x \right )\; }{x}[/latex]

 

The answer is 3, however I'm not sure how it's obtained. I know that [latex]\lim_{x\to 0}\frac{\; ln\left (1+x \right )\; }{x} = 1[/latex] but I can't quite figure out how to implement this property on the problem above. Could someone please give me a step-by-step explanation on how it should be solved? That would be great!

 

 

It seems like L'Hopital's Rule would suffice to solve this problem. Have you tried that?

It seems like L'Hopital's Rule would suffice to solve this problem. Have you tried that?

 

I haven't heard of it as far as I know. Am I obliged to use it or is there another way?

If you have proved that [latex]\lim_{x\to0}\frac{\ln(1+x)}x=1[/latex] (which can be done by L’Hôpital’s rule) then you can use it to deduce the given limit.

 

Hint: Let [latex]y=3x[/latex] and write the denominator in terms of [latex]y[/latex].

Edited March 25, 201312 yr by Nehushtan