Sunday, January 24, 2016

real analysis - Calculate limxto0fracln(1+2x)x2 with the help of l'Hospital's and Bernoullie's rule.




Task:


Calculate limx0ln(1+2x)x2

with the help of l'Hospital's and Bernoullie's rule.



My thoughts:


Because D(f)={xxRx0} the function is undefined for 0 and therefore, I need to find out, whether the function has a limit or only one-sided limits. In order to do that, I'll just calculate the one sided limits. If limx0+ln(1+2x)x2limx0ln(1+2x)x2limx0ln(1+2x)x2 doesn't exist


limx0ln(1+2x)x2lHospital==limx02/(2x+1)2x=limx01x(2x+1)productrule=limx01(2x+1)=1limx01x()=1()=()


(): If x is small, than 1/x gets proportional bigger. Let M>0 and let δ=1/M. Than $-1/x<\frac{-1}{1/M}=-M;\forall (-\delta). Since M can be arbitrarily large: limx01x=


limx0 analogue. limx0+ln(1+2x)x2==limx0+1x=

limx0 doesn't exist.


Is this proof correct?


Answer




Yes your evaluation is fine, to check it by standard limits, we have that


ln(1+2x)x2=ln(1+2x)2x2x1±


therefore the limit doesn't exist.


For the proof of the standard limit refer to


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