Trouble with Limits

[John45]

the first one is

Lim f(x) = (1/(5+h))-(1/5)) / h
x->0

and the second one is

Lim f(x) = (x^5-32) /(x-2)
x->2

it says to do this with the factorization formula which is (x^n-a^n)(x^(n-1)+(x^(n-2)a+x^(n-3)a^2+...xa^n-2+a^n-1) where n is a positive integer and a is a real number.

I'm having trouble of what to put for the "..."

any help would greatl be appreciated

 

[mmm4444bot]

Lim f(x) = (1/(5+h))-(1/5)) / h
x->0

Is your bold type in this thread supposed to signify something ?

You're missing an open parenthesis, somewhere.

The expression that defines function f above does not include the symbol x. Do you mean the limit as h approaches zero?

and the second one is

Lim f(x) = (x^5-32) /(x-2)
x->2

it says to do this with the factorization formula which is (x^n-a^n)(x^(n-1)+(x^(n-2)a+x^(n-3)a^2+...xa^n-2+a^n-1) where n is a positive integer and a is a real number.

Factor it by formula? Yuck.

I would simply divide x^5 - 32 by x - 2 using polynomial division, to get the other factor (i.e., the quotient):

x^4 + 2x^3 + 4x^2 + 8x + 16

If you would like to watch a short video on polynomial division, click HERE. Alternatively, you can google keywords polynomial division, for additional lessons on-line.

 

[galactus]

\(\displaystyle \lim_{h\to 0}\frac{\frac{1}{5+h}-\frac{1}{5}}{h}\)

Cross multiply and get:

\(\displaystyle \lim_{h\to 0}\frac{5-(h+5)}{5h(h+5)}\)

Now, simplify, cancel the h's that cancel and let the remaining h=0. See what you get.

\(\displaystyle \lim_{x\to 2}\frac{x^{5}-32}{x-2}\)

[/quote]

These expansions have a pattern. Note the exponents decrease by one and the coefficients are successive powers of 2.

\(\displaystyle x^{5}-(2)^{5}=(x-2)(x^{4}+2x^{3}+4x^{2}+8x+16)\)

Cancel the x-2 and plug in x=2 to find the limit.

As an aside, say we wanted to expand \(\displaystyle x^{10}-1024\)

This is \(\displaystyle x^{10}-(2)^{10}\)

By just using the pattern, we get \(\displaystyle (x-2)(x^{9}+2x^{8}+4x^{7}+8x^{6}+16x^{5}+32x^{4}+64x^{3}+128x^{2}+256x+512)\)

This factors some more if we wish, but that is enough to see the pattern.