Tuesday, 24 September 2019

calculus - A limit without invoking L'Hopital: limxto0fracxcosxsinxx2




The following limit



=limx0xcosxsinxx2



is a nice candidate for L'Hopital's Rule. This was given at a school before L'Hopital's Rule was covered. I wonder how we can skip the rule and use basic limits such as:



limx0sinxx,limx0cosx1x2


Answer



We have,




limx0xcosxsinxx2=limx0cosx1x+limx0xsinxx2



=2limx0sin2(x2)x+limx0xsinxx2



The first limit is zero since limx0sinxx=1, and,



0limx0xsinxx2limx0tanxsinxx2



But,




limx0tanxsinxx2=limx0 (sinx×1cosxx2cosx)=limx01cosxx=0



Thus, by the Squeeze Theorem,



limx0xcosxsinxx2=0


No comments:

Post a Comment

real analysis - How to find limhrightarrow0fracsin(ha)h

How to find limh0sin(ha)h without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...