Ytukyg

Recently I took a test where I was given these two limits to evaluate:
$lim_limits{h to 0}frac{sin(x+h)-sin{(x)}}{h}$ and $lim_limits{h to 0}frac{cos(x+h)-cos{(x)}}{h}.$ I used sine and cosine addition formulas and found the value of each limit individually, eventually canceling out $sin xcdot frac1h$ and $cos xcdot frac1h$ because I learned that we can evaluate limits piece by piece. As a result, I got $cos x $ and $-sin x$ as my two answers. However, my teacher marked it wrong saying that we cannot cancel $sin{x}cdotfrac1h$ or $cos{x}cdotfrac1h$ because those limits do not exist. Can someone please explain why this doesn't work? I thought that we can cancel those limits out since we never look at $0,$ just around $0,$ when evaluating these two limits.

Sine: $$frac{sin(x+h)-sin(x)}h=frac{sin(x)cos(h)+sin(h)cos(x)}h-frac{sin(x)}h$$

$$=sin(x)frac1h+cos(x)-sin(x)frac1h=cos(x)$$

Cosine: $$frac{cos(x+h)-cos(x)}h=frac{cos(x)cos(h)-sin(x)sin(h)}h-cos(x)frac1h$$

$$=cos(x)frac1h-sin(x)cdot1-cos(x)frac1h=-sin(x)$$

Note: I am allowed to assume $lim_limits{xto 0} frac{sin(h)}h=1,lim_limits{xto 0} frac{cos(h)-1}h=0.$

You broke up the limits incorrectly. That can be done only when the individual limits exist. $color{red}{lim_limits{h to 0} frac{sin x}{h}}$ and $color{red}{lim_limits{h to 0}frac{cos x}{h}}$ do NOT exist.

Here is how you can solve the first limit properly.

$$lim_{h to 0}frac{sin(x+h)-sin x}{h}$$

$$= lim_{h to 0}frac{color{blue}{sin xcos h}+cos xsin hcolor{blue}{-sin x}}{h}$$

$$= lim_{h to 0}frac{color{blue}{sin x(cos h-1)}+cos xsin h}{h}$$

$$= lim_{h to 0}frac{sin x(cos h-1)}{h}+lim_{h to 0}frac{cos xsin h}{h}$$

$$= sin xcdotlim_{h to 0}frac{cos h-1}{h}+cos xcdotlim_{h to 0}frac{sin h}{h}$$

Using $lim_limits{h to 0} frac{sin h}{h} = 1$ and $lim_limits{h to 0}frac{cos h-1}{h} = 0$, you get

$$= sin xcdot 0 + cos xcdot 1 = cos x$$

Notice how the individual limits exist. Therefore, the procedure is correct. Can you use the same way to solve for the second limit?

That's the standard way to find the derivatives for $sin x$ and $cos x$ by the definition.

Your way was not correct since as $hto 0$ we have $sin x/h not to 0$, to proceed properly for the first one we have

$$sin(x+h)-sin{(x)}=sin xcos h+sin hcos x-sin x$$

and then

$$lim_{h to 0}frac{sin(x+h)-sin{(x)}}{h}=lim_{h to 0}frac{sin xcos h+sin hcos x-sin x}{h}$$

$$=sin x cdotlim_{h to 0}frac{cos h-1}{h}+cos xlim_{h to 0}frac{sin h}{h}$$

and from here to conclude we need to use standard limits

• $lim_{h to 0}frac{1-cos h}{h^2}=frac12 implies lim_{h to 0}frac{1-cos h}{h}=0$


• $lim_{h to 0}frac{sin h}{h}=1$

For the other one we can proceed in a similar way.