Question

A^tanx-a^sinx/tanx-sinx when lim x yends to zero is equals to

Answer 1

it seems , you ask $\bold{\lim_{x\to\ 0}{\frac{a^{tanx}-a^{sinx}}{tanx-sinx}}}$
There are so many methods to solve it .
I suggest use standard form of limit .
I mean, $\bold{\lim_{f(x)\to\ 0}{\frac{a^{f(x)}-1}{f(x)}}}$ = loga
Use this standard form here ,

Now, $\bold{\lim_{f(x)\to\ 0}{\frac{(a^{tanx}-1)-(a^{sinx}-1)}{tanx-sinx}}}$
=$\bold{\lim_{f(x)\to\ 0}{\frac{\frac{(a^{tanx}-1)}{tanx}tanx-\frac{(a^{sinx}-1)}{sinx}sinx}{tanx-sinx}}}$
Now, apply above standard form ,
Then, $\bold{\lim_{f(x)\to\ 0}{\frac{\frac{(a^{tanx}-1)}{tanx}tanx-\frac{(a^{sinx}-1)}{sinx}sinx}{tanx-sinx}}}$ =( loga.tanx - loga.sinx)/(tanx - sinx)
= loga(tanx - sinx)/(tanx - sinx)
= loga

Hence, answer is loga