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/(1-3x) 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/(1-3x+h) - 2/(1-3x)]/h

I took the conjugate and came with

[4/(1-3x+h)^2 - 4/(1-3x)^2] / h(2/(1-3x+h) + 2/(1-3x))

Then I expanded the top and I got

(-8h - 24xh - 4h^2) / [(1-3x+h)^2 (1-3x)^2] / h(2/(1-3x+h) + 2/(1-3x))

I pulled out the "h" in the numerator and cancelled the "h" on the top and bottom and got

(-8 - 24x - 4) / [(1-3x+h)^2 (1-3x)^2] / (2/(1-3x+h) + 2/(1-3x))

I applied the limit

(-12 - 24x ) / [(1-3x)^4] / (4/(1-3x))

I get

(-12 + 24x) / 4(1-3x)^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/(1-3x+h) - 2/(1-3x)]/h

it should have been

[2/(1-3(x+h) ) - 2/(1-3x)]/h

= [2(1-3x) - 2(1 - 3x - 3h)]/((1-3x-3h)(1-3x)) * 1/h

= [2 - 6x - 2 + 6x + 6h]/((1-3x-3h)(1-3x)) * 1/h

= 6h/((1-3x-3h)(1-3x) * 1/h

= 6/((1-3x-3h)(1-3x))

so the limit of that as h ---> 0 = 6/(1-3x)^2

so when x = 0, the slope = 6/(1-0)^2 = 6

and when x = 0 , in the original y = 1/(1-0) = 1

so the y-intercept would be 1 and the slope = 6

equation: y = 6x + 1

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