Question on using the limit definition of the derivative to find f `(x)? I have tried looking at tutorials online and questions asked here but I cant wrap my head around it. Here is the question that I am stuck on.I am currently on part 1, I used the limit definition forumula but I ended up getting a fraction that I couldnt solve and I dont know where I went wrong. If anyone could help show me where I went wrong and how to do it I would really appreciate it!
 A: Using the definition of derivative we have
\begin{align}
\lim_{h\to 0}\dfrac{\dfrac{1}{x+h+2}-\dfrac{1}{x+2}}{h}&
\\ &=\lim_{h\to 0}\dfrac{\dfrac{x+2}{(x+2)(x+h+2)}-\dfrac{x+h+2}{(x+2)(x+h+2)}}{h}
\\&=\lim_{h\to 0}\dfrac{\dfrac{-h}{(x+2)(x+h+2)}}{h}
\\&=\lim_{h\to 0}\dfrac{-h}{h(x+2)(x+h+2)}
\\&=\lim_{h\to 0}\dfrac{-1}{(x+2)(x+h+2)}\\&=\dfrac{-1}{x+2)^2}.\end{align}
With respect to the third derivative we have
\begin{align}
\lim_{h\to 0}\dfrac{\cos(x+h)-\cos x}{h}&
\\ &=\lim_{h\to 0}\dfrac{\cos x\cos h-\sin x\sin h-\cos x}{h}
\\ &=\lim_{h\to 0}\dfrac{\cos x\cos h-\cos x}{h}-\lim_{h\to 0}\dfrac{\sin x\sin h}{h}
\\ &=\cos x\lim_{h\to 0}\dfrac{\cos h-1}{h}-\sin x\lim_{h\to 0}\dfrac{\sin h}{h}\\ &=0-\sin x=-\sin x.\end{align}
A: Part (iii).
$f(x)=\cos (x)$
$f(x+h)=\cos (x+h)=\cos x \cos h - \sin x \sin h$ (You need to know the compound angle formulae)
Then use small angle approximations: $\sin h \approx h$ and $\cos h \approx 1-\frac {h^2}2$
$f(x+h)-f(x)=\cos x \cos h - \sin x \sin h-\cos (x)$
$f(x+h)-f(x)=\cos x (1-\frac {h^2}2) - h\sin x -\cos (x)$
$f(x+h)-f(x)=\cos x -\frac {h^2}2 \cos x - h\sin x -\cos (x)$
$$\frac {f(x+h)-f(x)}h=\frac {\cos x -\frac {h^2}2 \cos x - h\sin x -\cos (x)}h$$
$$=\frac {-\frac {h^2}2 \cos x - h\sin x }h$$
$$=-\frac {h}2 \cos x - \sin x $$
$$\rightarrow -\sin x$$
