Question

How can we solve \[ \lim_{h \rightarrow 0}{ \frac{\|x+h\| - \|x\|}{h} } \]

Answer 1

h -> 0 but 0^- or 0^+??

Answer 2

just 0

Answer 3

The answer is \[ \|x\|/x \] because that is \[\frac{d}{dx} \|x\| \]
But not sure how I can get there with limits

Answer 4

0

Answer 5

It's not 0 the derivative on the real axis is defined in any point except for 0

Answer 6

lets use definition of |x|
|x| = sqrt(x^2)

making that substitution, we have

sqrt((x+h)^2) - sqrt(x^2) /h

on rationalizing ,
[ (x+h)^2 - x^2 ]/h (sqrt((x+h)^2) + sqrt(x^2))

(x+h -x)(x+h+x) /h (sqrt((x+h)^2) + sqrt(x^2))

=> (2x+h)/h (sqrt((x+h)^2) + sqrt(x^2))

now put h=0

=> 2x / 2sqrt(x^2)
=> x/|x|

hmm,,this was a good question..

Answer 7

Correction : in the 3rd last step ,
h won't be there in the denominator, it gets cancelled from the numerator.

Answer 8

how you get 2x+h all of a sudden?

Answer 9

@shubhamsrg

Answer 10

That's just the definition of the derivative for \(||x||\).

Answer 11

It is, but the question was how to solve it with limits

Answer 12

No it's |x|/x as I stated aboved and it's defined for x =/= 0 in R

Answer 13

@xcrypt Shrub showed you the steps. Did you get confused by any of that?

Answer 14

I don't know why he got 2x+h in the numerator all of a sudden

Answer 15

\[f(x)=|x|\]We're going to write it like this instead, \(f(x)=\sqrt{x^2}\)
So for \(f(x+h)\) we have, \(\sqrt{(x+h)^2}\)

So we'll plug these pieces into our derivative definition.\[f'(x)=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h} \qquad =\qquad \lim_{h \rightarrow 0}\frac{\sqrt{(x+h)^2}-\sqrt{x^2}}{h}\]This is the confusing part I guess :) You want to rationalize the top. So multiply the top and bottom by the CONJUGATE of the top. Understand how to do that?

Answer 16

\[\lim_{h \rightarrow 0}\frac{\sqrt{(x+h)^2}-\sqrt{x^2}}{h}\left(\frac{\sqrt{(x+h)^2}+\sqrt{x^2}}{\sqrt{(x+h)^2}+\sqrt{x^2}}\right)\]

Answer 17

The top will end up simplifying to 2xh+h^2, which will then further simplify to 2x+h.

If you're having trouble getting through the simplification let me know.

Answer 18

Well I used a^2 - b^2 = (a+b)(a-b) that will yield you (2x+h)(h)

also
|x|/x = x/|x|
both yield -1 when x<0
and +1 when x>0