The limit evaluation lim h→0 (f(2+h)-f(2))/h for the function f(x)=sin(x) is not equivalent to sin(2). The correct approach involves applying the definition of the derivative, which leads to the conclusion that the limit evaluates to cos(2). Numerical evaluation using small values of h, such as 1/2, 1/4, and 1/8, can provide insight into the behavior of the function near x=2, confirming that the limit approaches cos(2) rather than sin(2).
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No, the limit at the top is not the same as sin(2).AllanW said:Homework Statement
if f(x)=sin(x), evaluate lim h→0 (f(2+h)-f(2))/h) to two decimal places
Homework Equations
(f(x+h)-f(x))/h
The Attempt at a Solution
On the assumption that this is the same as lim x→2 sin(x)
sin(2)=0.91
AllanW said:Homework Statement
if f(x)=sin(x), evaluate lim h→0 (f(2+h)-f(2))/h) to two decimal places
Homework Equations
(f(x+h)-f(x))/h
The Attempt at a Solution
On the assumption that this is the same as lim x→2 sin(x)
sin(2)=0.91