Question

Limit help please!!
What is the limit of ((1/(x+ deltax)-(1/x))/deltax as deltax approaches 0 from the left?

Answer 1

Try simplifying the numerator.

Answer 2

I'm not sure how to.. I know when you subtract/add fractions you need a common denominator, but if you do that for the big numerator, do you also have to do it for the big denominator?

Answer 3

Not really. When you work on the numerator, ignore the denominator for the meantime. After simplifying the numerator, put the denominator back.

Answer 4

When I simplified the numerator and put the denominator back, I got delta x divided by (x + delta x) divided by delta x. Is that right so far?

Answer 5

\[\frac{\frac{\Delta x}{x+\Delta x}}{\Delta x}\ ?\]

Answer 6

yeah

Answer 7

That's not quite right.
\[\frac{1}{x+\Delta x}-\frac{1}{x}=\frac{x-(x+\Delta x)}{x(x+\Delta x)}\]

Answer 8

when i found the common denominator i multiplied the first fraction by 1 and the second fraction by (1+ delta x)

Answer 9

But \[\frac{1}{x}\cdot\frac{1}{1+\Delta x}=\frac{1}{x(1+\Delta x)}=\frac{1}{x+x\Delta x}\]
doesn't have the same denominator as \(\frac{1}{x+\Delta x}\).

Answer 10

Oh. I see what I did wrong, but what should I multiply it by?

Answer 11

You can multiply it by \(1/(x+\Delta x)\). Then you multiply the first factor by \(1/x\).

Answer 12

and when you simplify it the answer is zero?

Answer 13

Is it
\[\huge \lim_{h \rightarrow 0} \frac{\frac{1}{x+h}-\frac{1}{x}}{h}\]

Answer 14

No. It will be \[\frac{1}{x+\Delta x}-\frac{1}{x}=\frac{x-(x+\Delta x)}{x(x+\Delta x)}=\frac{-\Delta x}{x(x+\Delta x)}\]

Answer 15

Can you do it step by step? I'm not getting any of this

Answer 16

\[\begin{align}
\frac{1}{x+\Delta x}-\frac{1}{x}&=\frac{x}{x(x+\Delta x)}-\frac{x+\Delta x}{x(x+\Delta x)}\\
&=\frac{x-(x+\Delta x)}{x(x+\Delta x)}\\
&=\frac{x-x-\Delta x}{x(x+\Delta x)}\\
&=\frac{-\Delta x}{x(x+\Delta x)}\\
\end{align}\]

Answer 17

I think I'll just get my teacher to explain it - I'm just getting more confused :/