Question

How do you use the limit definition to find the slope of the tangent line to the graph `f(x)= 2x-x^2` at x=0?

Answer 1

The slope of the tangent when `x=0` is `2`

Explanation:

The slope of the tangent at any point is given by the derivative of the function at that point. The definition of the derivative of `y=f(x)` is

`f'(x)=lim_(h rarr 0) ( f(x+h)-f(x) ) / h`

So with `f(x) = 2x-x^2` and the value of `f'(0)` sought then;

`f'(0)=lim_(h rarr 0) (f(h)-f(0))/h`
`\ \ \ \ \ \ \ \ \=lim_(h rarr 0) ({2h-h^2}-{0-0})/h`

`\ \ \ \ \ \ \ \ \=lim_(h rarr 0) (2h-h^2)/h`

`\ \ \ \ \ \ \ \ \=lim_(h rarr 0) (2-h)`

`\ \ \ \ \ \ \ \ \=2`