# Reveiwing for test on limits

1. Oct 8, 2007

### 2slowtogofast

Im reveiwing for a test i have in about a week. Right now i have come to a problem i have a question on so here it is.

1. The problem statement, all variables and given/known data

lim x / x^2 - 1
x app infinity

3. The attempt at a solution

lim x / x^2 - 1
x app +infinity

x / x ( x - 1/x )

the x's cancel leaving

1 / (x-1/x)

giving

1 / +infinity - 1/+infinity

= 0+ ???
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 8, 2007

### NateTG

You should probably use parentheses to clarify things...
$$\lim_{x \rightarrow \infty} \frac{x}{x^2} -1=\lim_{x \rightarrow \infty} \frac{1}{x} - 1$$
$$\lim_{x \rightarrow \infty} \frac{x}{x^2-1}=\lim_{x \rightarrow \infty} \frac{1}{x-\frac{1}{x}}$$

Since $x-\frac{1}{x}$ and $x$ go to infinity, the fractions go to zero.

Last edited: Oct 8, 2007
3. Oct 8, 2007

### CompuChip

Agreed

How did you get this?
Did you use
$$\frac{1}{a + b} = \frac{1}{a} + \frac{1}{b}$$
by any chance? This is not true! For example,
$$\frac{1}{4} = \frac{1}{2 + 2} \stackrel{\large\mathbf{!}}{\not=} \frac{1}{2} + \frac{1}{2} = 1$$
Instead, if you have an expression of type
$$\lim_{x \to \infty} \frac{1}{f(x)}$$
with f(x) some function of x first try to determine the limit of f(x) as x goes to infinity. If f(x) goes to zero, the fraction goes to infinity (and vice versa) and if f(x) tends to a finite number F, then the limit of the fraction is 1/F.