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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 $\sin x\cdot \frac1h$ and $\cos x\cdot \frac1h$ 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 $\sin{x}\cdot\frac1h$ or $\cos{x}\cdot\frac1h$ 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 $\lim_\limits{x\to 0} \frac{\sin(h)}h=1,\lim_\limits{x\to 0} \frac{\cos(h)-1}h=0.$

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    $\begingroup$ Your answers are correct. (Those are the derivatives of $\sin x$ and $\cos x$.) Can you show your work? $\endgroup$ – KM101 Nov 24 '18 at 17:42
  • $\begingroup$ Show us your entire work. I cannot follow verbal descriptions. $\endgroup$ – Sean Roberson Nov 24 '18 at 17:43
  • $\begingroup$ You can cancel out $\frac{sin(h)}{h}$ when $h$ tends to $0$, but not $\frac{cos(h)}{h}$ as this limit tends to infinity $\endgroup$ – Sauhard Sharma Nov 24 '18 at 17:43
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    $\begingroup$ And by the way, you can only "cancel" when both limits exist. Note that $\lim_{h \to 0} 1/h$ does not exist. $\endgroup$ – Sean Roberson Nov 24 '18 at 17:44
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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.

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

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

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

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

$$= \sin x\cdot\lim_{h \to 0}\frac{\cos h-1}{h}+\cos x\cdot\lim_{h \to 0}\frac{\sin h}{h}$$

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

$$= \sin x\cdot 0 + \cos x\cdot 1 = \cos x$$

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

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  • $\begingroup$ Note that $\lim_\limits{h \to 0}\frac{\cos x}{h}=0$ for $\cos x =0$ then it is not always true that it doesn't exist. $\endgroup$ – user Nov 24 '18 at 18:11
  • $\begingroup$ Sorry for the nitpicking but it is not correct state that the limits doesn't exist at all the doesn't exist when $\sin x\neq 0$ and $\cos x \neq 0$. I know that it is a detail but I think it should be stated properly. $\endgroup$ – user Nov 25 '18 at 20:54
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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 $h\to 0$ we have $\sin x/h \not \to 0$, to proceed properly for the first one we have

$$\sin(x+h)-\sin{(x)}=\sin x\cos h+\sin h\cos x-\sin x$$

and then

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

$$=\sin x \cdot\lim_{h \to 0}\frac{\cos h-1}{h}+\cos x\lim_{h \to 0}\frac{\sin h}{h}$$

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.

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