Spart Posted March 23, 2013 Share Posted March 23, 2013 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! Link to comment Share on other sites More sharing options...
Cap'n Refsmmat Posted March 23, 2013 Share Posted March 23, 2013 It seems like L'Hopital's Rule would suffice to solve this problem. Have you tried that? Link to comment Share on other sites More sharing options...
Spart Posted March 23, 2013 Author Share Posted March 23, 2013 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? Link to comment Share on other sites More sharing options...
Nehushtan Posted March 25, 2013 Share Posted March 25, 2013 (edited) 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, 2013 by Nehushtan Link to comment Share on other sites More sharing options...
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