Limit Analysis: Simplifying Rational Expressions

[Shackleford]

1) = 0

The denominator grows much faster than the numerator.

2) = 0

Each of these terms would simply be zero, right?

For 4) and 5), the f(x) = x³

I simplified these to

4) = [(x-2)(x²+2x+4)]/(x-3)

5) = [(x-3)(x²+3x+9)]/(x-2)

http://i111.photobucket.com/albums/n149/camarolt4z28/Untitled2.png

http://i111.photobucket.com/albums/n149/camarolt4z28/Untitled.png

 

[HallsofIvy]

1) = 0

The denominator grows much faster than the numerator.

2) = 0

Each of these terms would simply be zero, right?

(1) is correct, (2) is not. In (2) your sequence is
$$\frac{1}{n}+ \frac{2}{n}+ \cdot\cdot\cdot+ \frac{n}{n^2}= \frac{1+ 2+ 3+ \cdot\cdot\cdot+ n}{n^2}$$

Use the fact that $1+ 2+ 3+ \cdot\cdot\cdot+ n= (1/2)n(n+1)$.

If you are thinking you can take the limit in each term so that you get 0 for each, no that is not correct.

For 4) and 5), the f(x) = x³

I simplified these to

4) = [(x-2)(x²+2x+4)]/(x-3)

You can do that but the obvious point should be that denominator goes to 0 while the numerator goes to f(3)- f(2)= 27- 8 which is NOT 0.

5) = [(x-3)(x²+3x+9)]/(x-2)

Again, the denominator goes to 0 while the numerator does not.

http://i111.photobucket.com/albums/n149/camarolt4z28/Untitled2.png

http://i111.photobucket.com/albums/n149/camarolt4z28/Untitled.png

 

[Shackleford]

For 2), I didn't know you could rewrite the sum in that manner.

I then get [(1/2)n(n+1)]/n = (1/2)[(n+1)/n] = 1/2.

If I had to guess, I would say 4) and 5) go to +/- infinity, respectively.

 

[Shackleford]

Am I right?