lim(sinx/x)m and lim 1-cosx /x
In this chapter, we complete what was mentioned briefly in the chapter on basic derivatives of trigonometric functions, about finding the derivative of the sine function.
To differentiate a function from first principles, we evaluate a limit. In the case of the function $\sin x$sinx where $x$x is a real number, the limit statement to be evaluated is
$\lim_{h\rightarrow0}\frac{\sin(x+h)-\sin x}{h}$limh→0sin(x+h)−sinxh
To make headway on this, we first expand the $\sin(x+h)$sin(x+h) term using a previously established identity. We have, $\sin(x+h)=\sin x\cos h+\cos x\sin h$sin(x+h)=sinxcosh+cosxsinh and therefore
| $\lim_{h\rightarrow0}\frac{\sin(x+h)-\sin x}{h}$limh→0sin(x+h)−sinxh | $=$= | $\lim_{h\rightarrow0}\frac{\sin x\cos h+\cos x\sin h-\sin x}{h}$limh→0sinxcosh+cosxsinh−sinxh |
| $=$= | $\lim_{h\rightarrow0}\frac{\sin x(\cos(h)-1)}{h}+\lim_{h\rightarrow0}\frac{\cos x\sin h}{h}$limh→0sinx(cos(h)−1)h+limh→0cosxsinhh |
We were justified in splitting the limit statement into two parts on the basis of another previously established fact about limits: that the limit of a sum is a sum of limits.
The problem would be solved if we knew how to evaluate the limits $\lim_{h\rightarrow0}\frac{\cos(h)-1}{h}$limh→0cos(h)−1hand $\lim_{h\rightarrow0}\frac{\sin h}{h}$limh→0sinhh. The following diagram makes this possible.
In the unit circle, the length of the vertical green line is $\sin h$sinh, the arc (in red) has length $h$h, and the vertical blue line is $\tan h$tanh. It is clear that for angles $h$h in the first quadrant, $\sin h\le h\le\tan h$sinh≤h≤tanh.
Dividing the inequalities by $\sin h$sinh, we have $1\le\frac{h}{\sin h}\le\frac{1}{\cos h}$1≤hsinh≤1cosh. Equivalently,
$1\ge\frac{\sin h}{h}\ge\cos h$1≥sinhh≥cosh.
Now, if we let $h\rightarrow0$h→0, then $\cos h\rightarrow1$cosh→1 and the fraction $\frac{\sin h}{h}$sinhh is between $1$1 and a number approaching $1$1. It must be that
$\lim_{h\rightarrow0}\frac{\sin h}{h}=1$limh→0sinhh=1
The limit, $\lim_{h\rightarrow0}\frac{\cos(h)-1}{h}$limh→0cos(h)−1h is found using the trick of multiplying the numerator and denominator by $\cos(h)+1$cos(h)+1. Thus,
| $\lim_{h\rightarrow0}\frac{\cos(h)-1}{h}$limh→0cos(h)−1h | $=$= | $\lim_{h\rightarrow0}\frac{\cos(h)-1}{h}.\frac{\cos(h)+1}{\cos(h)+1}$limh→0cos(h)−1h.cos(h)+1cos(h)+1 |
| $=$= | $\lim_{h\rightarrow0}\frac{\cos^2h-1}{h\left(\cos h+1\right)}$limh→0cos2h−1h(cosh+1) | |
| $=$= | $\lim_{h\rightarrow0}\frac{-\sin^2h}{h\left(\cos h+1\right)}$limh→0−sin2hh(cosh+1) | |
| $=$= | $\lim_{h\rightarrow0}\frac{-\sin h}{h}\times\lim_{h\rightarrow0}\frac{\sin h}{\cos h+1}$limh→0−sinhh×limh→0sinhcosh+1 |
Here, we again split the limit statement into two parts. This is justified by appealing to another previously established fact about limits: that the limit of a product is a product of limits.
The first limit statement on the right evaluates to $-1$−1 and the second evaluates to $0$0. Hence,
| $\lim_{h\rightarrow0}\frac{\cos(h)-1}{h}$limh→0cos(h)−1h | $=$= | $0$0 |
Putting these results together with an earlier stage of the discussion, we see that
| $\lim_{h\rightarrow0}\frac{\sin(x+h)-\sin x}{h}$limh→0sin(x+h)−sinxh | $=$= | $\cos x$cosx |
That is, the derivative of the sine function is the cosine function,
We can find $\frac{\mathrm{d}}{\mathrm{d}x}\cos x$ddxcosx by first writing $\cos x$cosx as $\sin(\frac{\pi}{2}-x)$sin(π2−x). Using the result just obtained and the function-of-a-function rule, we have
| $\frac{\mathrm{d}}{\mathrm{d}x}\cos x$ddxcosx | $=$= | $\frac{\mathrm{d}}{\mathrm{d}x}\sin(\frac{\pi}{2}-x)$ddxsin(π2−x) |
| $=$= | $-\cos(\frac{\pi}{2}-x)$−cos(π2−x) | |
| $=$= | $-\sin x$−sinx |
Example 1
Evaluate $\lim_{x\rightarrow0}\frac{\sin(3x)}{2x}$limx→0sin(3x)2x.
We know that $\lim_{x\rightarrow0}\frac{\sin x}{x}=1$limx→0sinxx=1. So, $\lim_{x\rightarrow0}\frac{\sin(3x)}{3x}=1$limx→0sin(3x)3x=1. The denominator in the question will be correct if we rewrite the limit as $\lim_{x\rightarrow0}\frac{3}{2}\frac{\sin(3x)}{3x},$limx→032sin(3x)3x, which is the same as $\frac{3}{2}\times\lim_{x\rightarrow0}\frac{\sin(3x)}{3x}$32×limx→0sin(3x)3x. (The constant multiple can be brought outside the limit sign.) Thus,
$\lim_{x\rightarrow0}\frac{\sin(3x)}{2x}=\frac{3}{2}$limx→0sin(3x)2x=32.
Example 2
Evaluate $\lim_{x\rightarrow0}\frac{\sin(3x)}{\sin(2x)}$limx→0sin(3x)sin(2x).
We might make use of the fact that $x$x and $\sin x$sinx are almost identical for small values of $x$x. Thus,
| $\lim_{x\rightarrow0}\frac{\sin(3x)}{\sin(2x)}$limx→0sin(3x)sin(2x) | $\approx$≈ | $\lim_{x\rightarrow0}\frac{\sin(3x)}{2x}$limx→0sin(3x)2x |
| $=$= | $\frac{3}{2}\lim_{x\rightarrow0}\frac{\sin(3x)}{3x}$32limx→0sin(3x)3x | |
| $=$= | $\frac{3}{2}$32 |
Another approach would be to use multiple angle formulae to express $\sin(3x)$sin(3x) as $3\sin x\cos^2x-\sin^3x$3sinxcos2x−sin3x and $\sin(2x)$sin(2x) as $2\sin x\cos x$2sinxcosx. Then, the fraction simplifies to $\frac{4\cos^2x-1}{2\cos x}$4cos2x−12cosx. This clearly has the limit $\frac{3}{2}$32 as $x\rightarrow0$x→0.
Example 3
Evaluate $\lim_{x\rightarrow0}\frac{1+3\sin x-\cos x}{2x}$limx→01+3sinx−cosx2x.
We split the limit statement into manageable pieces. It is the same as
$\lim_{x\rightarrow0}\frac{1-\cos x}{2x}+\lim_{x\rightarrow0}\frac{3\sin x}{2x}$limx→01−cosx2x+limx→03sinx2x.
This is,$-\frac{1}{2}\lim_{x\rightarrow0}\frac{\cos x-1}{x}+\frac{3}{2}\frac{\sin x}{x}$−12limx→0cosx−1x+32sinxx which evaluates to $\frac{3}{2}$32.
More Examples
Question 1
Evaluate $\lim_{x\to0}\left(\frac{7\sin x}{x}\right)$limx→0(7sinxx).
Question 2
Evaluate $\lim_{x\to0}\left(\frac{\sin7x}{7x}\right)$limx→0(sin7x7x).
Question 3
Evaluate $\lim_{x\to0}\left(\frac{\sin4x}{\sin3x}\right)$limx→0(sin4xsin3x).