Question

How do i go foward with this

Answer 1

$$\lim_{\Delta x\to 0^-}\frac{\frac{1}{x+\Delta x}-\frac{1}{x}}{x}=\lim_{\Delta x\to 0^-}\frac{\frac{x}{x}\cdot \frac{1}{x+\Delta x}-\frac{1}{x}\cdot\frac{x+\Delta x}{x+\Delta x}}{x}=\lim_{\Delta x\to 0^-}\frac{\frac{x}{x(x+\Delta x)}-\frac{x+\Delta x}{x(x+\Delta x)}}{x}=$$
$$=\lim_{\Delta x\to 0^-}\frac{\frac{-\Delta x}{x(x+\Delta x)}}{x}=\lim_{\Delta x\to 0^-}\frac{-\Delta x}{x(x+\Delta x)}\cdot \frac{1}{x}$$

Answer 2

are you sure, the denominator isn't
\(\Delta h \)
?

Answer 3

oh yes it is -.- delta x

Answer 4

so then it becomes 1/(x(x+delta(x)))

Answer 5

yep, now just plug in
\(\Delta h =0\)

Answer 6

\(\Delta x = 0\)

Answer 7

also there should be a negative sign

Answer 8

but then what about the "X"

Answer 9

keep x as x
the answer would be in terms of x

Answer 10

1/(x(x+0))

Answer 11

where is the negative sign ?

Answer 12

1/(x(x+0^-))

Answer 13

no, i meant this:
\(\Large \dfrac{-\Delta x}{x(x+\Delta x)}\cdot \frac{1}{\Delta x}\)

Answer 14

ohh -(1)/(x(x+0))

Answer 15

yesssss
thats
\(\Large \dfrac{-1}{x^2}\)

Answer 16

got it thanks

Answer 17

but then how would i type this question through wolfram?

Answer 18

take \(\Delta x =h\)
and then try

Answer 19

oh so instead of typing delta(x) i would use simply h?

Answer 20

got thanks

Answer 21

yes and welcome ^_^