Solving Limits: How to Find the Limit of a Function as x Approaches Infinity

[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.

Homework Statement

lim x / x^2 - 1
x app infinity

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+ ?

 

[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.

 

[CompuChip]

the x's cancel leaving

1 / (x-1/x)

Agreed

giving

1 / +infinity - 1/+infinity

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.