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14. Evaluate the following limits  \lim_{x \rightarrow 0} \frac{\sin ax }{\sin bx } , a,b \neq 0

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The limit,

\lim_{x \rightarrow 0} \frac{\sin ax }{\sin bx } , a,b \neq 0

On putting the limit directly, the function takes the zero-by-zero form. So, we convert it in the form of \frac{sina}{a} and then put the limit,

\Rightarrow \lim_{x\rightarrow {0}}\frac{\frac{sinax}{ax}}{\frac{sinbx}{bx}}\cdot\frac{ax}{bx}

=\frac{\lim_{ax\rightarrow {0}}\frac{sinax}{ax}}{\lim_{bx\rightarrow {0}}\frac{sinbx}{bx}}\cdot\frac{a}{b}

=\frac{a}{b}   (Answer)

Posted by

Pankaj Sanodiya

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