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

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SUMMARY

The limit of the function as x approaches infinity for the expression lim (x / (x^2 - 1)) is determined to be 0. The discussion highlights the simplification process where x cancels out, leading to the expression 1 / (x - 1/x). As x approaches infinity, both x and 1/x tend to infinity, resulting in the limit converging to 0. The importance of correctly applying limit properties, particularly the behavior of fractions as functions approach infinity, is emphasized.

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2slowtogofast
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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+ ?
 
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You should probably use parentheses to clarify things...
[tex]\lim_{x \rightarrow \infty} \frac{x}{x^2} -1=\lim_{x \rightarrow \infty} \frac{1}{x} - 1[/tex]
[tex]\lim_{x \rightarrow \infty} \frac{x}{x^2-1}=\lim_{x \rightarrow \infty} \frac{1}{x-\frac{1}{x}}[/tex]

Since [itex]x-\frac{1}{x}[/itex] and [itex]x[/itex] go to infinity, the fractions go to zero.
 
Last edited:
2slowtogofast said:
the x's cancel leaving

1 / (x-1/x)
Agreed

2slowtogofast said:
giving

1 / +infinity - 1/+infinity
How did you get this?
Did you use
[tex]\frac{1}{a + b} = \frac{1}{a} + \frac{1}{b}[/tex]
by any chance? This is not true! For example,
[tex]\frac{1}{4} = \frac{1}{2 + 2} \stackrel{\large\mathbf{!}}{\not=} \frac{1}{2} + \frac{1}{2} = 1[/tex]
Instead, if you have an expression of type
[tex]\lim_{x \to \infty} \frac{1}{f(x)}[/tex]
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.
 

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