The limit represents the derivative of some role f(x) at some number a. Uncover f and a.

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lim (h->0) of ((7+h)^2-49)/h

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You need to discover the derivative the the role at x=a=7, hence, using the limit an interpretation 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...

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Suppose the the role is `f(x)=x^2` and also `a = 7` .

You need to discover the derivative the the role at x=a=7, hence, utilizing the limit meaning 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 choose 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 also `((7+h)^2-49)/h` yields x + h = 7 + h and `x^2 = 49` .

Notice that het relations provides `x_(1,2) = +-7` however the an initial relation x + h = 7 + h excludes the worth -7, hence x = 7.

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

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