Question

How do you evaluate `(sin(2h)(1 - cos h)) / h` as h approaches 0?

Answer 1

`lim_(h->0) (sin(2h)(1-cos(h)))/(h) = 0`

Explanation:

Because direct substitution of `h = 0` into this expression yields and indeterminate form of `0/0`, we can evaluate this limit using L'Hospital's rule.

In the case of the indeterminate form `0/0`, we take the derivative of the numerator and denominator.

Thus, we can write

`lim_(h->0) (sin(2h)(1-cos(h)))/(h) = lim_(h->0) (sin(2h)-sin(2h)cos(h))/(h)`

`= lim_(h->0) (d/(dh)(sin(2h)) - d/(dh)(sin(2h)cos(h)))/(d/(dh)(h))`

`= lim_(h->0) (2cos(2h) - (2cos(2h)cos(h)+sin(h)*-sin(h)))/(1)`

`= lim_(h->0) (2cos(2h)(1-cos(h))+sin(2h)sin(h) = 0`

By graphing our original function, we can see that it does in fact equal to `0`.

graph{(sin(2x)(1-cos(x)))/(x) [-1.922, 1.923, -0.963, 0.96]}

Answer 2

The reqd. limit `=0`.

Explanation:

We use the Identity `sin2h=2sinhcoh` and the fact

`lim_(hraa0)sinh/h=1`

The reqd. limit`=lim_(hrarr0)2(sinh/h)cosh(1-cosh)=2*1*cos0*(1-cos0)`

`=2*1*1*(1-1)=0`.