# Finding Limits

• Sep 11th 2008, 03:04 PM
stones44
Finding Limits
Ok so I'm used to using graphs to find limits, but I can't use a calculator for this..I need to learn the equation way of doing it
1. lim as x--> 8 x^2 - 9x + 8 / x^2 - 10x +16
im bad at simplifying algebra so if someone could step by step with explanations that would help..i realize that as it approaches 8 the equation turns into 0/0 but what does that mean in terms of its limit

2. lim as x --> 3+ x^2 + 2x + 1 / x^2 - 2x - 3
i just dont get how to do it without graphs...

EDIT:
3x^3 - 12x^2 + 9x / 2x^3 - 12x^2 + 18x as x --> 0

same: as x --> +infinity

same: as x --> 3

and i can use my calc for this one below..am i right?
lim as x--> 3 (2^x) - 8 / x-3 is it +infinity
• Sep 11th 2008, 03:26 PM
silencecloak
Quote:

Originally Posted by stones44
Ok so I'm used to using graphs to find limits, but I can't use a calculator for this..I need to learn the equation way of doing it
1. lim as x--> 8 x^2 - 9x + 8 / x^2 - 10x +16
im bad at simplifying algebra so if someone could step by step with explanations that would help..i realize that as it approaches 8 the equation turns into 0/0 but what does that mean in terms of its limit

2. lim as x --> 3+ x^2 + 2x + 1 / x^2 - 2x - 3
i just dont get how to do it without graphs...

$\displaystyle \frac{(x-8)(x-1)}{(x-8)(x-2)} = \frac{(x-1)}{(x-2)} = \frac{(8-1)}{(8-2)} = \frac{7}{6}$
• Sep 11th 2008, 03:42 PM
stones44
OK so i need to factor....ok

well i factored the second and i still have a 0 at the denom..also, without using a graph how do i tell the from the right and from the left apart

also, for the first one, the graph shows that the left and right sides dont meet..so doesnt that mean there is no limit?
• Sep 11th 2008, 03:50 PM
silencecloak
Quote:

Originally Posted by stones44
OK so i need to factor....ok

well i factored the second and i still have a 0 at the denom..also, without using a graph how do i tell the from the right and from the left apart

also, for the first one, the graph shows that the left and right sides dont meet..so doesnt that mean there is no limit?

well I am no expert at limits but I would say the 2nd equation has no limit since the denominator = 0

and the first ones limit is the 7/6

don't rely on the graph in my opinion
• Sep 11th 2008, 03:56 PM
stones44
ok thanks

and another harder one:

3x^3 - 12x^2 + 9x / 2x^3 - 12x^2 + 18x as x --> 0

same: as x --> +infinity

same: as x --> 3
• Sep 13th 2008, 03:04 AM
Moo
Hello,
Quote:

Originally Posted by stones44
ok thanks

and another harder one:

3x^3 - 12x^2 + 9x / 2x^3 - 12x^2 + 18x as x --> 0

same: as x --> +infinity

same: as x --> 3

Factor out 3x in the numerator and 2x in the denominator.

$\displaystyle =\frac{3x}{2x} \cdot \frac{x^2-4x+3}{x^2-6x+9}=\frac 32 \cdot \frac{(x-3)(x-1)}{(x-3)^2}=\frac 32 \cdot \frac{x-1}{x-3}$

It is defined in x=0

As $\displaystyle x \to \infty$, divide the numerator and the denominator by x : $\displaystyle \frac 32 \cdot \frac{1-\frac 1x}{1-\frac 3x}$

As $\displaystyle x \to 3$, differentiate the two situations : $\displaystyle x \to 3^+$ and $\displaystyle x \to 3^-$