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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#

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]}

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Answer 2

Scarlett Allison

The reqd. limit #=0#.

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#.

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Answer 3

Samantha Adkison

The limit of (sin(2h)(1 - cos h)) / h as h approaches 0 is equal to 2.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.