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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