Task:
Calculate limx→0ln(1+2x)x2
with the help of l'Hospital's and Bernoullie's rule.
My thoughts:
Because D(f)={x∣x∈R∧x≠0} 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 limx→0+ln(1+2x)x2≠limx→0−ln(1+2x)x2⟹limx→0ln(1+2x)x2 doesn't exist
limx→0−ln(1+2x)x2l′Hospital⏞==limx→0−2/(2x+1)2x=limx→0−1x(2x+1)product−rule⏞=limx→0−1(2x+1)⏟=1⋅limx→0−1x⏟(∗)=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)
limx→0− analogue. limx→0+ln(1+2x)x2=⋯=limx→0+1x=∞
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)2x2x→1⋅±∞
therefore the limit doesn't exist.
For the proof of the standard limit refer to
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