Posted by Daniel on Saturday, October 1, 2011 at 7:23pm.
I am trying to find equation of the tangent line to the curve 2/(13x) at x=0 using the definition.
I applied it into the formula
lim h>0 [f(a+h)  f(a)]/h
which gave me
[2/(13x+h)  2/(13x)]/h
I took the conjugate and came with
[4/(13x+h)^2  4/(13x)^2] / h(2/(13x+h) + 2/(13x))
Then I expanded the top and I got
(8h  24xh  4h^2) / [(13x+h)^2 (13x)^2] / h(2/(13x+h) + 2/(13x))
I pulled out the "h" in the numerator and cancelled the "h" on the top and bottom and got
(8  24x  4) / [(13x+h)^2 (13x)^2] / (2/(13x+h) + 2/(13x))
I applied the limit
(12  24x ) / [(13x)^4] / (4/(13x))
I get
(12 + 24x) / 4(13x)^3
From what I have done, it looks like my work is right, but the answer I come up with
y=3x + 2
However, the answer is y=6x + 2
I can not see what I did wrong.

calculus  Reiny, Saturday, October 1, 2011 at 10:48pm
from your
[2/(13x+h)  2/(13x)]/h
it should have been
[2/(13(x+h) )  2/(13x)]/h
= [2(13x)  2(1  3x  3h)]/((13x3h)(13x)) * 1/h
= [2  6x  2 + 6x + 6h]/((13x3h)(13x)) * 1/h
= 6h/((13x3h)(13x) * 1/h
= 6/((13x3h)(13x))
so the limit of that as h > 0 = 6/(13x)^2
so when x = 0, the slope = 6/(10)^2 = 6
and when x = 0 , in the original y = 1/(10) = 1
so the yintercept would be 1 and the slope = 6
equation: y = 6x + 1
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