# Compute this limit $\lim_{x\to0}\frac{\sin(x^2+\frac{1}{x})-\sin\frac{1}{x}}{x}$ using L'Hôpital's rule

I have asked this problem before, but I can't understand the explanation, I couldn't understand how the sin multiply for cos, and too multiply for A + and - B: $$\sin(A)-\sin(B)=2\sin\left(\frac{A-B}{2}\right)\cos\left(\frac{A+B}{2}\right)$$ and I don't understand in this step how/why the $$A-B$$ and $$A+B$$ was replaced by $$\frac{x^2}{2}$$ and $$\frac{x^2}{2}+\frac{1}{x}$$ : $$\lim_{x\to0}\frac{\sin\left(x^2+\frac1x\right)-\sin\left(\frac1x\right)}{x}= \lim_{x\to0}\frac{2\sin\left(\frac{x^2}{2}\right)\cos\left(\frac{x^2}{2}+\frac1x\right)}{x}.$$

• If you have asked this problem before, maybe you could give the link to that previous question? – celtschk Jun 19 '13 at 18:18
• @user73276:i cant use L'hôpital's rule because $\lim_{x\to0}{\sin(x^2+\frac{1}{x})-\sin\frac{1}{x}}\neq0$ – M.H Jun 19 '13 at 18:45

Hint:its easy to prove $$\sin(x+y)+\sin(x-y)=2\sin(x)\cos(y)$$ then put $y=\frac{A+B}{2}$,$x=\frac{A-B}{2}$ $$\sin(x)\sim x$$ $$\ cosx\sim1-\frac{x^2}{2}$$ because $$\lim_{x\to0}\frac{sinx}{x}=\lim_{x\to0}\frac{cosx}{1-\frac{x^2}{2}}=1$$

It's easier if you use the plain old addition formula:

$$\sin{\left(x^2+\frac{1}{x}\right)} = \cos{x^2}\, \sin{\frac{1}{x}} + \sin{x^2} \,\cos{\frac{1}{x}}$$

which behaves as

$$\left(1-\frac12 x^2\right) \sin{\frac{1}{x}} + x^2 \cos{\frac{1}{x}}$$

so that

$$\frac{\sin{\left(x^2+\frac{1}{x}\right)} - \sin{\frac{1}{x}}}{x} \sim \frac{x^2 \cos{\frac{1}{x}} - (1/2)x^2 \sin{\frac{1}{x}}}{x}$$

which goes to zero as $x \to 0$.