# help with limits...and secant lines

• Jan 28th 2010, 12:43 PM
ryan18
help with limits...and secant lines
Use a table of values to estimate the value of the limit. If you have a graphing device, use it to confirm your result graphically.
http://www.webassign.net/cgi-bin/sym...5E4%20-%201%29

When I graph this, as it is going to 1 it looks like it reaches and passes 0.5...

---------------------

The point P(1, 1/9) lies on the curve y = x/(8 + x). If Q is the point (x, x/(8 + x)), use a scientific calculator to find the slope of the secant line PQ (correct to six decimal places) for the following values of x.

(a) 0.5
(b) 0.9
(c) 0.99
(d) 0.999
(e) 1.5
(f) 1.1
(g) 1.01
(h) 1.001

If you could just help me with how to solve these I should be able to work them out myself.
• Jan 28th 2010, 12:54 PM
Dinkydoe
$\displaystyle \lim_{x\to 1}\frac{x^6-1}{x^4-1} = \lim_{x\to 1}\frac{(x-1)(x^5+x^4+x^3+x^2+x+1)}{(x-1)(x^3+x^2+x+1)}$ =

$\displaystyle \lim_{x\to 1}\frac{x^5+x^4+x^3+x^2+x+1}{x^3+x^2+x+1}$

So now you can fill in $\displaystyle x=1$
• Jan 28th 2010, 01:10 PM
Dinkydoe
For 2. I don't see why you'd want to use a calculator if you can calculate the exact equation with the derivative $\displaystyle y'(x)$.

The derivative is given by $\displaystyle y'(x) = \frac{1}{8+x}-\frac{x}{(8+x)^2}= \frac{8}{(8+x)^2}$

Then the tangent line at $\displaystyle x=p$ is given by $\displaystyle y'(p)(x-p)+y(p)$
• Jan 28th 2010, 01:34 PM
ryan18
For the limits equation you gave me, if I plugged in x=1 then it would give me $\displaystyle 5/4$ and that isnt the correct answer...

And I dont really understand what you mean in the second problem you gave me.
• Jan 28th 2010, 02:03 PM
Dinkydoe
It's a typo: after the "=" sign I forgot to put an extra "x" in the numerator.
It was supposed to be: $\displaystyle \lim_{x\to 1}\frac{x^5+x^4+x^3+x^2+x+1}{x^3+x^2+x+1}$

Now plug in $\displaystyle x= 1$. It gives $\displaystyle \frac{3}{2}$

L'hopital -Theorem confirms this:

$\displaystyle \lim_{x\to 1}\frac{x^6-1}{x^4-1} = \lim_{x\to 1}\frac{6x^5}{4x^3} = \frac{6}{4} = \frac{3}{2}$

Second problem: The derivative as I earlier stated gives a formula for the slope of the point: $\displaystyle (x,y(x))$ at any point $\displaystyle x$.

$\displaystyle y'(x) = \frac{8}{(8+x)^2}$. So if you want to know the slope at the point $\displaystyle (1,y(1))$ the derivative gives: $\displaystyle y'(1) = \frac{8}{81}$ and the tangent at $\displaystyle (1,y(1))$ is given by

$\displaystyle f(x) = \frac{8}{81}(x-1)+ \frac{1}{9}$
• Jan 28th 2010, 02:16 PM
ryan18
ok so for the second problem should I just plug in any number I want to find?
Just plug in each number and find the result?
(a) 0.5
(b) 0.9
(c) 0.99
(d) 0.999
(e) 1.5
(f) 1.1
(g) 1.01
(h) 1.001
• Jan 28th 2010, 02:19 PM
Dinkydoe
Exactly:

$\displaystyle y'(x) = \frac{8}{(8+x)^2}$

So fill in your values of $\displaystyle x$