Monday, 9 June 2014

limits - Finding limxto0fracasinbxbsinaxx2sinax witouth L'Hopital, what is my mistake?



I was working on this question.



lim



\lim_{x \to 0} \dfrac {1}{x^2} \cdot \lim_{x \to 0} \dfrac { \frac {1}{abx}}{\frac {1}{abx}} \cdot \dfrac {a\sin bx -b\sin ax}{\sin ax}




\lim_{x \to 0} \dfrac {1}{x^2} \cdot \dfrac {\lim_{x \to 0} \frac {\sin bx}{bx}- \lim_{x \to 0}\frac{\sin ax}{ax}}{\frac 1b \lim_{x \to 0} \frac {\sin ax}{ax}}



b \lim_{x \to 0} \dfrac {1}{x^2} \cdot \dfrac {b-a}{a}



It seems like this limit does not exist, but if you apply L'Hopital's rule you seem to get an answer. What is wrong with what I did?


Answer



One error is that
\lim_{x \to 0} \frac{\sin cx}{cx} = 1, not c,
so your fraction is

\frac1{x^2}\frac{1-1}{1/b}
which is still indeterminate.


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