Finding the limit

1. The problem statement, all variables and given/known data

find the limit if it exist :
lim (1/[x+Δχ] - 1/x) / Δχ
x->0-





3. The attempt at a solution
lim (1/[x+Δχ] - 1/x) / Δχ => lim -Δχ /(x+ Δx) / Δχ
x->0-
then what ? do i cancel the ΔX in the numerator with that in the denominator ? or whats the next step to solve this problem and did I do any mistake?
 

Dick

Science Advisor
Homework Helper
26,249
611
That limit is undefined as x->0. Are you sure you don't mean delta X ->0? And, if so, sure, you want to cancel the thing that's going to 0. And don't write things like a/b/c without parentheses. It's not clear whether you mean (a/b)/c or a/(b/c). They are different.
 
oh yes, I'm sorry , I miss typed it its supposed to be delta x approaching 0 from the left hand side of the graph.
 

Dick

Science Advisor
Homework Helper
26,249
611
Ok. Let's write h instead of delta X, ok. You want limit h->0 (1/(x+h)-1/x)/h. I'm not really happy with -h/(x+h)/h for reasons beyond the parentheses.
 
460
0
You have 1/(x+dx) - 1/x

Cross multiply them, then see what you're left with.

edit: Oh you already did that, the dx cancels and as dx->0 it tends to -1/x^2 no?
 
it matches my answer but i wanted to make sure that i did the proper steps
 
31,930
3,893
You have 1/(x+dx) - 1/x
Cross multiply them, then see what you're left with.
I think I understand what you meant to say, but cross multiplication applies when you have an equation with two rational expressions, such as
a/b = c/d

"Cross multiplication" results in ad = bc, and is equivalent to multiplying both sides of the equation by bd.
 
460
0
I think I understand what you meant to say, but cross multiplication applies when you have an equation with two rational expressions, such as
a/b = c/d

"Cross multiplication" results in ad = bc, and is equivalent to multiplying both sides of the equation by bd.
I mean

x/x(x+dx) - (x+dx)/x(x+dx) to get -dx/(x^2+dx)
 

HallsofIvy

Science Advisor
41,626
821
He meant, "get a common denominator and subtract the two fractions."
 

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