# Confusion on the given tutorial.

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Hey guys,

based on 2nd post in the link

$\frac{(1+h)^2 - 1}{h} = \frac{1 + 2h + h^2 - 1}{h} = \frac{2h + h^2}{h} = \frac{h(2 + h)}{h} = 2 + h$

we can cut and all of that if x = 1.

but how about if we express the answer in terms of x?

$\frac{(x+h)^2 - x}{h} = \frac{x^2 + 2xh + 2h - x}{h} = ?$

seems like quadratic equation, doesn't it will produce 2 answer? (where we only get one gradient(tangent) only, right?)

btw, it's still not available to be answered because the h can't be 0.

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You just made a tiny mistake, you should be looking at:

$\lim_{h \to 0} \frac{(x+h)^2 - x^2}{h}$

you'll find it works out easily.

Edited by the tree
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Hey guys,

now, I'm having problem get some intuitive on Chain Rule.

Chain Rule Proof

where it said:

$\frac{g(x + h) - g(x)}{h} - g'(x) = v \rightarrow 0$, as $h \rightarrow 0$

but then, when I tried to:

$\frac{g(x + 0) - g(x)}{0} - g'(x) = \frac{g(x) - g(x)}{0} - 1 = \frac{0}{0} - 1 = 0?$

doesn't $\frac{0}{0}$ = undefined?

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Taking the limit as h -> 0 is different to just substituting in h=0.

You'll need a better idea of limits before trying to work your way through that proof.

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• 3 weeks later...

You are taking a limit so you can avoid $\frac{0}{0}$.

Remember $h \rightarrow 0$ means that $h$ is approaching (or tends to) $0$ and should not be simply substituted in.

What you must remember is that you are trying to find the gradient of a tangent slope at a single point.

The general idea behind what you are doing is: You need to make a secant line between two different points, if they were the same point you could not find the slope as you would have $\frac{0}{0}$. This however does not give us an accurate tangent slope. To overcome this problem you let the distance between those two points approach 0 (so the slope of the secant line gets ever closer to the tangent slope you are looking for), and that is where taking the limit helps us.

I suggest, as did the tree; go back and learn limits fully before trying to tackle the chain rule.

Edited by Pixel

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