# Math Help - Finding tangent slopes using derivative limit definitions

1. ## Finding tangent slopes using derivative limit definitions

Hello, I'm required to use the definition:

m = lim as x approaches a of (f(x)-f(a))/(x-a)

to find the slope of the tangent line of the function: y=4x-x^2 at the point (1,3).

So I now have:

lim x->1 [(4x-x^2)-f(1)]/(x-1) which comes out to
lim x->1 (4x-x^2-3)/(x-1).

I still have 0 in the denominator for direct substitution so I thought about multiplying by the conjugate x+1 but that would still give me x^2-1 which would still be 0. Any thoughts on how to continue this problem?

Thanks.

2. factorise numerator and x-1 gets cancelled out.put the limit and ur answer comes out.

3. Originally Posted by buddyp450
Hello,
to find the slope of the tangent line of the function: y=4x-x^2 at the point (1,3).
lim x->1 [(4x-x^2)-f(1)]/(x-1) which comes out to
lim x->1 (4x-x^2-3)/(x-1).

I still have 0 in the denominator for direct substitution so I thought about multiplying by the conjugate x+1 but that would still give me x^2-1 which would still be 0. Any thoughts on how to continue this problem?
Note that $4x-x^2-3=(x-1)(-x+3)$.

4. Thanks again! I'm so bad at factoring. >.<

...since it wasn't in the form of the x^2 first, the 4x second and the number last I didn't recognize it... duh.

5. Originally Posted by buddyp450
Thanks again! I'm so bad at factoring. >.<

...since it wasn't in the form of the x^2 first, the 4x second and the number last I didn't recognize it... duh.
Here's something that will help you with these kinds of problems:
The limit of $\frac{f(x)- f(a)}{x-a}$, as x goes to a, always involves an "indeterminant" of the form 0/0. In particular, f(a)- f(a)= 0. If f is a polynomial, that means that f(x)- f(a) has a factor of x-a. Since the whole point is to be able to cancel the "x-a" in the denominator, always try that- divide the numerator by x-a to find the other factor.