1. ## limits

Consider the function $\frac{x}{{1 - |x|}}$

1) Give a reason why the quotient rule for differentiation cannot be used to find g'(0)
2) Determine whether or not g is differentiable at x=0 by evaluating separately the left and right hand limits:
$\mathop {\lim }\limits_{h \to {0^ - }} \frac{{g(h) - g(0)}}{h}$ and $\mathop {\lim }\limits_{h \to {0^ + }} \frac{{g(h) - g(0)}}{h}$
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3) Why is it that derivatives do not exist at end points?

2. 1) Does the absolute-value part have a derivative everywhere? (Think about where the graph of y = |x| has an "elbow"....)

2) For this part, just find the limits. Since |x| = -x for x < 0 and |x| = +x for x > 0, the limits will not have the same sign.

3) Can you have two one-sided limits in agreement if you have only one one-sided limit?

3. It appears you got the definition of the Left hand derivative wrong.It goes like this

2) $g_{+}'(0)=\lim_{h\to 0}\frac{g(0+h)-g(0)}{h}=\lim_{h\to 0}\frac{g(h)}{h}=\lim_{h\to 0}\frac{1}{1-|h|}=1$

$g_{-}'(0)=\lim_{h\to 0}\frac{g(0-h)-g(0)}{-h}=\lim_{h\to 0}\frac{g(-h)}{-h}=\lim_{h\to 0}\frac{1}{1-|-h|}=1$

So you see the derivative indeed exists.

Using quotient rule we may get
$
g'(x)=\frac{(1-|x|).1-x(-\frac{|x|}{x})}{(1-|x|)^2}=\frac{1-|x|+|x|}{(1-|x|)^2}=\frac{1}{(1-|x|)^2}
$

So you still get the derivative at $x=0$ as $1$ but there is a slight hitch since
$\frac{d}{dx}(|x|)=\frac{|x|}{x}$
would not exist at $x=0$,though by some kind of default we do get our answer.

Let us see what other members have to say