Originally Posted by

**theowne** I have two questions, both are kind of similar involving rationalization and I would appreciate if someone can push me in the correct direction.

The first is finding the limit of this:

__sqrt(3x-5) - 4__

x-7

And this is as limit x-> 7

I tried to rationalize and got this in the end

__3(x-7)__

(x-7)(sqrt(3x-5) - 4) Mr F says: That - should be a +.

But after canceling out the (x-7), the denominator is still 0. I'm not sure where I can go from there, I think I must have made a mistake somewhere..

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The second question is specifically finding the slope to the tangent at a given point from first principles.

of y = sqrt{2x+8} at (-2,2)

So this was the form I used,

Limit h->0

__sqrt{2(2+h)+8} - 2__

h

Mr F says: x = -2 so the bit in red should be -2. So you need to re-work the solution from here.

[snip]