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India
Class XI

lim(sinx/x)m and lim 1-cosx /x

Lesson

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 

$\lim_{h\rightarrow0}\frac{\sin(x+h)-\sin x}{h}$limh0sin(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

$\lim_{h\rightarrow0}\frac{\sin(x+h)-\sin x}{h}$limh0sin(x+h)sinxh $=$= $\lim_{h\rightarrow0}\frac{\sin x\cos h+\cos x\sin h-\sin x}{h}$limh0sinxcosh+cosxsinhsinxh
  $=$= $\lim_{h\rightarrow0}\frac{\sin x(\cos(h)-1)}{h}+\lim_{h\rightarrow0}\frac{\cos x\sin h}{h}$limh0sinx(cos(h)1)h+limh0cosxsinhh

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 $\lim_{h\rightarrow0}\frac{\cos(h)-1}{h}$limh0cos(h)1hand $\lim_{h\rightarrow0}\frac{\sin h}{h}$limh0sinhh. 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, $\sin h\le h\le\tan h$sinhhtanh.

Dividing the inequalities by $\sin h$sinh, we have $1\le\frac{h}{\sin h}\le\frac{1}{\cos h}$1hsinh1cosh. Equivalently, 

$1\ge\frac{\sin h}{h}\ge\cos h$1sinhhcosh.

Now, if we let $h\rightarrow0$h0, then $\cos h\rightarrow1$cosh1 and the fraction $\frac{\sin h}{h}$sinhh is between $1$1 and a number approaching $1$1. It must be that 

$\lim_{h\rightarrow0}\frac{\sin h}{h}=1$limh0sinhh=1

 

The limit, $\lim_{h\rightarrow0}\frac{\cos(h)-1}{h}$limh0cos(h)1h is found using the trick of multiplying the numerator and denominator by $\cos(h)+1$cos(h)+1. Thus, 

$\lim_{h\rightarrow0}\frac{\cos(h)-1}{h}$limh0cos(h)1h $=$= $\lim_{h\rightarrow0}\frac{\cos(h)-1}{h}.\frac{\cos(h)+1}{\cos(h)+1}$limh0cos(h)1h.cos(h)+1cos(h)+1
  $=$= $\lim_{h\rightarrow0}\frac{\cos^2h-1}{h\left(\cos h+1\right)}$limh0cos2h1h(cosh+1)
  $=$= $\lim_{h\rightarrow0}\frac{-\sin^2h}{h\left(\cos h+1\right)}$limh0sin2hh(cosh+1)
  $=$= $\lim_{h\rightarrow0}\frac{-\sin h}{h}\times\lim_{h\rightarrow0}\frac{\sin h}{\cos h+1}$limh0sinhh×limh0sinhcosh+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, 

$\lim_{h\rightarrow0}\frac{\cos(h)-1}{h}$limh0cos(h)1h $=$= $0$0

Putting these results together with an earlier stage of the discussion, we see that 

$\lim_{h\rightarrow0}\frac{\sin(x+h)-\sin x}{h}$limh0sin(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(π2x). 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(π2x)
  $=$= $-\cos(\frac{\pi}{2}-x)$cos(π2x)
  $=$= $-\sin x$sinx

 

Example 1

Evaluate $\lim_{x\rightarrow0}\frac{\sin(3x)}{2x}$limx0sin(3x)2x.

We know that $\lim_{x\rightarrow0}\frac{\sin x}{x}=1$limx0sinxx=1. So, $\lim_{x\rightarrow0}\frac{\sin(3x)}{3x}=1$limx0sin(3x)3x=1. The denominator in the question will be correct if we rewrite the limit as $\lim_{x\rightarrow0}\frac{3}{2}\frac{\sin(3x)}{3x},$limx032sin(3x)3x, which is the same as $\frac{3}{2}\times\lim_{x\rightarrow0}\frac{\sin(3x)}{3x}$32×limx0sin(3x)3x.  (The constant multiple can be brought outside the limit sign.) Thus, 

$\lim_{x\rightarrow0}\frac{\sin(3x)}{2x}=\frac{3}{2}$limx0sin(3x)2x=32.

 

Example 2

Evaluate $\lim_{x\rightarrow0}\frac{\sin(3x)}{\sin(2x)}$limx0sin(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, 

$\lim_{x\rightarrow0}\frac{\sin(3x)}{\sin(2x)}$limx0sin(3x)sin(2x) $\approx$ $\lim_{x\rightarrow0}\frac{\sin(3x)}{2x}$limx0sin(3x)2x
  $=$= $\frac{3}{2}\lim_{x\rightarrow0}\frac{\sin(3x)}{3x}$32limx0sin(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$3sinxcos2xsin3x 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}$4cos2x12cosx. This clearly has the limit $\frac{3}{2}$32 as $x\rightarrow0$x0.

 

Example 3

Evaluate  $\lim_{x\rightarrow0}\frac{1+3\sin x-\cos x}{2x}$limx01+3sinxcosx2x.

We split the limit statement into manageable pieces. It is the same as

$\lim_{x\rightarrow0}\frac{1-\cos x}{2x}+\lim_{x\rightarrow0}\frac{3\sin x}{2x}$limx01cosx2x+limx03sinx2x

This is,$-\frac{1}{2}\lim_{x\rightarrow0}\frac{\cos x-1}{x}+\frac{3}{2}\frac{\sin x}{x}$12limx0cosx1x+32sinxx which evaluates to $\frac{3}{2}$32.

 

More Examples

Question 1

Evaluate $\lim_{x\to0}\left(\frac{7\sin x}{x}\right)$limx0(7sinxx).

Question 2

Evaluate $\lim_{x\to0}\left(\frac{\sin7x}{7x}\right)$limx0(sin7x7x).

Question 3

Evaluate $\lim_{x\to0}\left(\frac{\sin4x}{\sin3x}\right)$limx0(sin4xsin3x).

Outcomes

11.C.LD.1

Derivative introduced as rate of change both as that of distance function and geometrically, intuitive idea of limit. Definition of derivative, relate it to slope of tangent of the curve, derivative of sum, difference, product and quotient of functions. Derivatives of polynomial and trigonometric functions.

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