Evaluate the limit of $(f(2+h)-f(2))/h$ as $h$ approaches $0$ for $f(x) = \sin(x)$.

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!

4 Answers

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.

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)

"[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.

Use Prosthaphaeresis Formulas,

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

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