If f(x) = sinx, evaluate f(2+h)-f(2)/h to two decimal places as h approaches 0.

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amansadaiqy · Accepted Answer

Given: L=limh→0sin⁡(2+h)−sin⁡(2)hL=limh→0sin⁡2cos⁡h+cos⁡2sin⁡h−sin⁡2hL=limh→0sin⁡2cos⁡h−sin⁡2h+cos⁡2sin⁡hhL=limh→0sin⁡2(cos⁡h−1h)+cos⁡2sin⁡hhL=limh→0sin⁡2(cos⁡h−1h)+limh→0cos⁡2sin⁡hhWe now employ the well-known limits of calculus limθ→0sin⁡θθ=1 and limβ→0cos⁡β−1β=0 to evaluate. L=sin⁡2(0)+cos⁡2(1)L=cos⁡(2)Which is roughly equivalent to L≈−0.42And I assumed that cos⁡(2) was in radians.

Sage Farley · Accepted Answer

As an alternative, we can employ the derivative's limit definition:

f^{'} (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h}

This exactly matches our limit if you plug in x=2, so what we're looking for then is:

f^{'} (2)

Since

f (x) = \sin x

, we know that the derivative of f(x) is:

f^{'} (x) = \cos x

Then our limit equals

lim_{h \to 0} \frac{f (2 + h) - f (2)}{h} = f^{'} (2) = \cos (2) \approx - 0.42

If f(x) = sinx, evaluate f(2+h)-f(2)/h to two decimal places as h approaches 0.

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