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# for x is not equal to 0, what is the limit of 1/h[ (1/ x + h) - (1/x) ] when h approaches 0 - Kathryn T. from Houston, TX

Hello Kathryn,

I'm guessing you mean to put the second h in the denominator with the first x. Then we'll have limh→0 1/h[(1/(x+h) - (1/x)] = limh→0 1/h[-h/(x(x+h))] = limh→0 -1/(x(x+h) and when we substitute 0 for h we get -1/x2, final answer.

Hmmm, is the question find the limit of 1/x as x approaches zero and you have to use the formula limit of the limit of [f(0+h)-f(0-h)]/[(0+h)-(0-h)] as h approaches zero.  This is a really difficult concept to learn as a beginner to calculus and one that often makes a lot more sense when you start learning what you are actually measuring is the same thing as a slope.

The formula [f(x+h)-f(x)]/[(x+h)-(x)] is actually just another version of the (y1-y2)/(x2-x1)

I mean, you can find the way to solve this one from the example above, but if you don't know what you are actually doing, it won't do you any good. You might want to search khan academy and introduction to limits and they have 3 or 4 5 minute videos which will help you with understanding what you are doing.

I might be reading your question wrong though, and if I am I apologize.

This limit is the definition of derivatives with f(x) = 1/x plugged in so it is f'(x) = -1/x^2

You need to add the fractions together:

(1/h){[1/(x+h)] - [1/x]} = (1/h){-h/[x(x+h)]} = -1/[x(x+h)]

Now you can apply the limit to the function above:

lim(h-->0) -1/[x(x+h)] = -1/(x^2)