Sorry! I don't get this one either..
limit of (f(x+Δx)-f(x)) / Δ x with Δ x -> 0
f(x)= 1/(x+3)
My work so far before I got stuck...
1/((x+Δx)+3)
= 1/(x+Δx+3) - 1/x+3
= 1(x+3)/(x+Δx+3)(x+3) - 1(x+Δx+3)/(x+3)(x+Δx+3)
=Δx / (x+Δx+3)(x+3)
I need to put Δx/(x+Δx+3)(x+3) divided by Δx, but I don't know how to solve that..
oh, much more fun !
ALMOST !!! sign error
- dx /[(x+dx+3)(x+3) ]
divide by dx
-1/[(x+dx+3)(x+3) ]
now let dx -->0
-1/(x+3)^2
which any calculus text will tell you is the right answer
d/dx (1/u) = [bottom*d top - top d bottom]/bottom squared
OH opps! Thank you!!
To find the limit of (f(x+Δx)-f(x)) / Δx as Δx approaches 0, you can simplify the expression further.
You correctly found the expression 1/((x+Δx)+3) for f(x+Δx). Now, let's find f(x):
f(x) = 1/(x+3)
Next, substitute these values into the expression (f(x+Δx)-f(x)) / Δx:
(f(x+Δx)-f(x)) / Δx = [1/(x+Δx+3) - 1/(x+3)] / Δx
Now, we can combine the terms over a common denominator:
= [(x+3) - (x+Δx+3)] / [(x+3)(x+Δx+3)Δx]
= (x+3 - x - Δx - 3) / [(x+3)(x+Δx+3)Δx]
= -Δx / [(x+3)(x+Δx+3)Δx]
The Δx terms cancel out:
= -1 / [(x+3)(x+Δx+3)]
Now, let's simplify this further. As Δx approaches 0, we can substitute 0 for Δx:
= -1 / [(x+3)(x+0+3)]
= -1 / [(x+3)(x+3)]
So, the final simplified expression is -1 / [(x+3)(x+3)].
Now, to find the limit as Δx approaches 0, you can substitute 0 for Δx in the simplified expression:
lim Δx→0 (-1 / [(x+3)(x+3)])
And that is the limit of the given expression as Δx approaches 0.