If f(x) = sinx, evaluate f(2+h)-f(2)/h to two decimal places as h approaches 0. Help?!
2 Answers
The limit equals
Explanation:
We have
#L = lim_(h->0) (sin(2 + h) - sin(2))/h#
#L = lim_(h->0) (sin2cos h + cos2sin h - sin2)/h#
#L = lim_(h->0) (sin2cos h - sin2)/h +(cos2sin h)/h#
#L = lim_(h->0) sin2((cos h - 1)/h) + cos2(sin h)/h#
#L = lim_(h->0) sin2((cos h -1 )/h) + lim_(h->0) cos2(sin h)/h#
We now use the well known calculus limits of
#L = sin2(0) + cos2(1)#
#L = cos(2)#
Which can be approximated to
#L ~~ -0.42#
And I assumed that
Hopefully this helps!
Explanation:
Alternatively, we can use the limit definition of the derivative:
#f'(x) = lim_(h->0)(f(x+h)-f(x))/h#
This exactly matches our limit if you plug in
#f'(2)#
Since
#f'(x) = cosx#
Therefore our limit equals
#lim_(h->0)(f(2+h)-f(2))/h = f'(2) = cos(2) ~~ -0.42#
Final Answer