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

2 Answers
Jul 19, 2016

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]}

Jul 20, 2016

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.