Question

lim of (x tends to y) (sinx-siny) divided by (x-y)...
Please help me find the solution.. & with brief.

Answer 1

Answer:

The answer is $\cos x$

Step-by-step explanation:

$\displaystyle \lim_{x\to y}\frac{\sin x-\sin y}{x-y}=\displaystyle \lim_{x\to y}\frac{2\sin\frac{x-y}{2}\cos\frac{x+y}{2}}{x-y}\\=\displaystyle \lim_{x\to y}\frac{\sin\frac{x-y}{2}}{\frac{x-y}{2}}(\cos\frac{x+y}{2})\\=1(\cos\frac{x+x}{2})=\cos x, \because \displaystyle \lim_{x\to 0}\frac{\sin x}{x}=1$