# The limit represents the derivative of some function f(x) at some number a. Find f and a. lim (h->0) of ((7+h)^2-49)/h PLEASE HELP

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Suppose that the function is `f(x)=x^2` and `a = 7` .

You need to find the derivative of the function at x=a=7, hence, using the limit definition of derivatives yields:

`f'(x) = lim_(h-gt0) (f(x+h)-f(x))/h`

`f'(x) = lim_(h-gt0) ((x+h)^2-x^2)/h`

Expanding the binomial yields:

`f'(x) = lim_(h-gt0) (x^2 + 2xh + h^2 - x^2)/h`

Reducing like terms yields:

`f'(x) = lim_(h-gt0) (2xh + h^2)/h`

Factoring out h yields:

`f'(x) = lim_(h-gt0) h(2x + h)/h =gt f'(x) = lim_(h-gt0) (2x + h) = 2x`

Equating `((x+h)^2-x^2)/h` and `((7+h)^2-49)/h` yields x + h = 7 + h and `x^2 = 49` .

Notice that het relations gives `x_(1,2) = +-7` but the first relation x + h = 7 + h excludes the value -7, hence x = 7.

**Hence, evaluating the function yields that `f(x) = x^2` and a=7 => f'(7) = 14.**