
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]