If $f(x)=\sin x$, evaluate $\displaystyle\lim_{h \to 0} \frac{f(2+h) - f(2)}{h}$ to two decimal places.

Have tried to answer in many different ways, but always end up getting confused along the way, and am unable to cancel terms.

Any help is always appreciated!

  • $\begingroup$ You should look in your book at the definition of the derivative of a function $\endgroup$ – charlestoncrabb Nov 7 '15 at 23:37
  • $\begingroup$ It's equal to $\sin'(2)$. $\endgroup$ – user236182 Nov 7 '15 at 23:37
  • $\begingroup$ It's very hard to give an answer to your question before you've told us if you know what a derivative is and if you know the derivative of $\sin(x)$. If your answer is no to both questions, then the answer given by Benedict Voltaire below is for you. If you answered yes to both questions, then you could recognize this limit as being the derivative of $\sin(x)$ when $x=2$. $\endgroup$ – Bernard Massé Nov 9 '15 at 15:27

Your problem wants you to evaluate

$$\lim_{h \to 0} \frac{f(2+h) - f(2)}{h}$$

where $f(x) = \sin(x)$ and $x=2$.

First, note that $f(2+h) = \sin(2+h) = \sin(2)\cos(h) + \sin(h)\cos(2)$, which is a trig identity.

Then, the expression, $f(2+h) - f(2)$ simply becomes $\sin(2)\cos(h) + \sin(h)\cos(2) - \sin(2)$.

Therefore, we have to evaluate the following limit

$$\lim_{h \to 0} \frac{\sin(2)\cos(h) + \sin(h)\cos(2) - \sin(2)}{h} \tag{1}$$

We want to be able to use the following trig limit identities in our problem

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

Thus, we will rewrite $(1)$ into the following form

$$\frac{\sin(2)\cos(h) + \sin(h)\cos(2) - \sin(2)}{h} = \sin(2)\cdot\frac{(\cos(h)-1)}{h} + \cos(2) \cdot \frac{\sin(h)}{h} $$

Thus evaluating the limit, we deduce

$$\lim_{h \to 0} \frac{\sin(2)\cos(h) + \sin(h)\cos(2) - \sin(2)}{h} = \lim_{h \to 0} \bigg( \sin(2)\cdot\frac{(\cos(h)-1)}{h} \bigg) + \lim_{h \to 0} \bigg( \cos(2) \cdot \frac{\sin(h)}{h} \bigg)$$

which becomes

$$\sin(2) \cdot 0 + \cos(2) \cdot 1 = \cos(2)$$

as desired.

Using a calculator, we see that $\cos(2) \approx -0.42$ radians.

When you learn derivatives, you’ll find that $\frac{\mathrm d}{\mathrm dx}(\sin(x)) = \cos(x)$, and when $x = 2$, then $\frac{\mathrm d}{\mathrm dx}(\sin(2)) = \cos(2)$, which we have formally proved above using the definition of the derivative.

  • $\begingroup$ While this is all true, this is a little redundant.... why not just give a link to a solution for the derivative of $\sin x$?? It's better to just prove it in general than to solve what is effectively the same problem over and over again. Extremely detailed answer nevertheless $\endgroup$ – Brevan Ellefsen Nov 8 '15 at 0:12

Remember that $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ Here we see that $f(x) = \sin x$ and $x = 2$. Therefore, all we have to do is find $f'(2)$. If you know that the derivative of the sine function is $\cos x$ then you immediately have the answer as $\cos (2)$ (Remember that this is in Radians... all of Calculus should be done in radians)

  • $\begingroup$ One problem with this is that there must be some way to find the limit independently of knowing the derivative if one is to use the limit to prove that $\sin'=\cos$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Nov 7 '15 at 23:56
  • $\begingroup$ There is a way... it just mirrors the proof that $\sin' = \cos$. You'll end up with something very close to the classical $\frac{\sin x}{x}$ in the proof that will require further limits that are very hard to show without first knowing the derivative of sine... in short, it's much easier to just look at a proof of $\sin'(x)$ using limits and use that the way I do above $\endgroup$ – Brevan Ellefsen Nov 8 '15 at 0:04

"[I] am unable to cancel terms."

This suggests you're trying to do some algebra. But the phrase "to two decimal places" suggests that doing arithmetic was intended instead. The means you must somehow find $\sin 2$ and $\sin2.01$ or some other value of the sine for a number near $2$. If we assume calculators are infallible, we get $$ \frac{\sin(2+0.01) - \sin 2}{0.01} \approx \frac{0.9050906 - 0.9092974}{0.01} \approx -0.4206864. $$ Since $\cos2\approx-0.4161468$, this is within $0.01$ of the right value.

Remember: This is in radians, not in degrees.

This limit is usually found not by doing cancelations, but by squeezing.

  • 1
    $\begingroup$ While I like your answer, I would point out that asking for a specific amount of digits does not necessarily imply arithmetic was intended... this is probably for a class and the teacher needs a way to check answers, so having one student provide $\cos 2$ to one decimal place and another student to 20 places would be very hard to check; as a result the number of reported digits would be predetermined $\endgroup$ – Brevan Ellefsen Nov 8 '15 at 0:07

Use Prosthaphaeresis Formulas,

$$\sin(a+h)-\sin a=2\sin\dfrac h2\cos\dfrac{2a+h}2$$

$$\lim_{h\to0}\dfrac{\sin(a+h)-\sin a}h=\lim_{h\to0}\dfrac{\sin\dfrac h2}{\dfrac h2}\cdot\lim_{h\to0}\cos\dfrac{2a+h}2=?$$


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