Question

How do you find the limit of `(3 sin (x)(1 - cos(x))) / (x^2)` as x approaches 0?

Answer 1

`Lt_(x->0)(3sin(x)(1-cos(x)))/x^2=0`

Explanation:

To find `Lt_(x->0)(3sin(x)(1-cos(x)))/x^2`, let us divide it in two parts

`Lt_(x->0)(3sin(x))/x xx Lt_(x->0)(1-cos(x))/x`

Now as each of these is of the form `0/0` when `x->0`,

we can use L'Hospital's Rule,

and as such `Lt_(x->0)(3sin(x))/x xx Lt_(x->0)(1-cos(x))/x`

= `Lt_(x->0)(d/(dx)3sin(x))/1 xx Lt_(x->0)(d/(dx)(1-cos(x)))/1`

= `Lt_(x->0)(3cos(x))/1 xx Lt_(x->0)(0-(-sin(x)))/1`

= `3xx0/1=0`
graph{(3sin(x)(1-cos(x)))/x^2 [-2.5, 2.5, -1.25, 1.25]}

Answer 2

`0`

Explanation:

`1-cos(x)=cos(0)-cos(x)`

we know that

`cos(a+b)-cos(a-b)=-2sin(a)sin(b)`

calling

`{(a+b=0),(a-b=x):}`

and solving for `a,b`

we have `a=x/2,b=-x/2` so

`cos(0)-cos(x)=2sin^2(x/2)`

then

`(3sin(x)(1-cos(x)))/x^2 = (6sin(x)sin^2(x/2))/x^2`

Now using the fundamental result

`lim_(x->0)sin(x)/x=1` we have

`lim_(x->0)(3sin(x)(1-cos(x)))/x^2=6/4lim_(x->0)sin(x)lim_(x->0)(sin(x/2)/(x/2))^2 = 3/2 cdot 0 cdot 1 = 0`

Answer 3

We can use the two fundamental trigonometric limits.

Explanation:

We have

`lim_(xrarr0)sinx/x = 1` `" "` and `" "` `lim_(xrarr0)(1-cosx)/x = 0`.

These limits, together with the product property of limits allow us to write:

`lim_(xrarr0)(3sinx(1-cosx))/x^2 = 3lim_(xrarr0)(sinx/x) lim_(xrarr0) (1-cosx)/x`

`= 3(1)(0) = 0`