# Limit on Mars.

• Oct 7th 2010, 04:12 PM
DannyMath
Limit on Mars.
I've managed to find a solution to the following question using the f(a+h) - f(a) / h definition of the tangent line limit, but I got caught up trying to get the same answer using the: lim x -> 1 f(x) - f(1) / x-1 definition:

If a rock is thrown upward on the planet Mars with a velocity of 10 m/s, its height (in meters) after t seconds is given by H = 10t - 1.86t^2

(a) find the velocity of the rock after one second (I got 6.28 m/s)
(b) find the velocity of the rock when t = a (I got 10-3.72a)

The question doesn't ask about the second definition but I'm just curious to know what mistake I made while trying to calculate it using the f(x) - f(1) definition. Thanks for any clarification :)
• Oct 7th 2010, 05:20 PM
skeeter
might be easier to identify a mistake if you show your attempt ...
• Oct 7th 2010, 05:58 PM
DannyMath
Hey skeeter, thanks for the reply! I was hesitant to show my work (some people can be harsh on forums). From what I can read of my scribble-heavy writing, this was the last step I got:

lim x -> 1 (10x - 1.86x^2 - (10 - 1.86))/ x-1

then

lim x -> 1 (-1.86x^2 + 10x - 8.14 )/ x-1

I tried to factor out a -1.86 hoping to be able to work with what was in the brackets to get an "x-1" term, but this is the last thing I have in my notes...

lim x -> 1 -1.86(x^2 - (10/1.86)x + (8.14/1.86)) / x-1
• Oct 8th 2010, 05:47 AM
HallsofIvy
Basic rule of algebra: If, for polynomial p(x), p(a)= 0, then x- a is a factor of p(x).

Note that when x= 1, f(x)- f(1)= f(1)- f(1)= 0.

Specifically, when x= 1, 10(1)- 1.86(1^2)- (10- 1.86)= 10- 1.86- 10+ 1.86= 0

so that -186(1^2)+ 10(1)- 8.14= 0 which tells you that x- 1 is a factor of -1.86x^2+ 10x- 8.14! Divide that quadratic by x- 1 to find the other factor, P(x). Once you know that, your limit becomes $\displaystyle \lim_{x\to 1} \frac{(x-1)P(x)}{x-1}= \lim_{x\to 1}P(x)$.