can someone evaluate this for me?

If you take this limit as h->0, this is a derivative problem. Using the product rule, you know that:

d[x^-2+2]/dx = -2/x^3. Now check if that's what we get for the limit.

You are given f(x). To evaluate f(x+h), plug in x+h instead of x. Thus, the abouve expression simplifies to:

[(1/(x+h)^2 + 2) - (1/x^2 + 2)]/h ----> simplifying, you get:

[1/(x^2+2xh+h^2) - 1/x^2]/h -----> multiply by the common denominator to simplify:

{[x^2 - (x^2+2xh+h^2)]/[(x^2)(x^2+2xh+h^2)]}/h ----> simplfy to get:

(-2x-h)/(x^4+2x^3h+x^2h^2)

This is your desired answer, but if you are looking for the derivative, evaluate this as h->0. Doing this, you get:

-2/x^3, which is the derivative from above

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Hope this helps

I assume you're looking for the derivative of 1/x^2+2, which is just -2/x^3.

ok:

[1/(x+h)^2 +2 - (1/x^2 +2 ) ]/h

= [1/(x+h)^2 - 1/x^2 ] / h

f(x+h)-f(x)/h when f(x) = 1/x^2 + 2

[1/(x+h)^2 +2 - (1/x^2 +2 ) ]/h

= [1/(x+h)^2 - 1/x^2 ] / h

= [x^2 -(x+h)^2]/ x^2(x+h)^2 h

= [x^2 -x^2 -2xh - h^2]/ x^2(x+h)^2 h

= [-2xh - h^2]/ x^2(x+h)^2 h

= h [-2x - h]/ x^2(x+h)^2 h

= [-2x - h]/ x^2(x+h)^2

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