Lesson

If the value of a function approaches some finite quantity as the domain variable becomes arbitrarily large, then that function value is loosely called a 'limit at infinity'. Such a function may have a graph that looks something like the following.

A function $f(x)$`f`(`x`) that approaches a finite limit as $x$`x` becomes large, is said to have a *horizontal asymptote*.

A function may have an asymptote that is not horizontal. In such a case, the function does not approach a finite limit. That is, there is no limit at infinity. This can be seen by observing that such an asymptote is either vertical or it is a line with the form $y(x)=ax+b$`y`(`x`)=`a``x`+`b` where a $\ne0$≠0, which is clearly unbounded as $x\rightarrow\infty$`x`→∞.

Certain rational functions have horizontal asymptotes. For this to occur, the denominator must be a polynomial of degree equal to or higher than the degree of the numerator polynomial.

To investigate the asymptotic behaviour of rational functions, a useful fact to remember is the limit$\lim_{x\rightarrow\infty}\frac{1}{x}=0$lim`x`→∞1`x`=0. This is often used in combination with the limit laws discussed in another chapter.

In cases involving trigonometric functions, it can be helpful to recall the limits $\lim_{x\rightarrow0}\frac{\sin x}{x}=1$lim`x`→0`s``i``n``x``x`=1 and $\lim_{x\rightarrow0}\frac{\cos x-1}{x}=0$lim`x`→0`c``o``s``x`−1`x`=0.

Find the limit, if it exists, of the function $x\sin\frac{1}{x}$`x``s``i``n`1`x` as $x\rightarrow\infty$`x`→∞.

As $x$`x` becomes arbitrarily large, the expression approaches the form $\infty\times0$∞×0 which is indeterminate. However, $x\sin\frac{1}{x}$`x``s``i``n`1`x` can be rewritten as $\frac{\sin\frac{1}{x}}{\frac{1}{x}}$`s``i``n`1`x`1`x`.

We know that $\lim_{y\rightarrow0}\frac{\sin y}{y}=1$lim`y`→0`s``i``n``y``y`=1 and $\lim_{x\rightarrow\infty}\frac{1}{x}=0$lim`x`→∞1`x`=0. Now, if we put $y=\frac{1}{x}$`y`=1`x`, we have $\lim_{x\rightarrow\infty}\frac{\sin\frac{1}{x}}{\frac{1}{x}}=\lim_{y\rightarrow0}\frac{\sin y}{y}=1$lim`x`→∞`s``i``n`1`x`1`x`=lim`y`→0`s``i``n``y``y`=1.

Investigate whether there are limits as $x\rightarrow-\infty$`x`→−∞ and $x\rightarrow\infty$`x`→∞ for the function $y(x)=\frac{x^3-2x}{2x^3+x^2-1}$`y`(`x`)=`x`3−2`x`2`x`3+`x`2−1.

We rewrite the function as $\frac{1-\frac{2}{x^2}}{2+\frac{1}{x}-\frac{1}{x^3}}$1−2`x`22+1`x`−1`x`3 by dividing the numerator and denominator by $x^3$`x`3. Now, because $\lim_{x\rightarrow\infty}\frac{1}{x}=0$lim`x`→∞1`x`=0 it must be true (by a limit theorem) that $\lim_{x\rightarrow\infty}\frac{1}{x^2}=0$lim`x`→∞1`x`2=0 and $\lim_{x\rightarrow\infty}\frac{1}{x^3}=0$lim`x`→∞1`x`3=0. It will not make any difference whether $x$`x` approaches positive or negative infinity.

Therefore, the limit we seek is $\frac{1}{2}$12.

Find the value of $\lim_{x\to\infty}\left(x\sin\left(\frac{4}{x}\right)\right)$lim`x`→∞(`x``s``i``n`(4`x`)).

Find the value of $\lim_{x\to\infty}\left(x-\sqrt{x^2+7}\right)$lim`x`→∞(`x`−√`x`2+7).

Use the graph of $y=2-e^{-x}$`y`=2−`e`−`x` to find the value of $\lim_{x\to\infty}\left(2-e^{-x}\right)$lim`x`→∞(2−`e`−`x`).

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Identify discontinuities and limits of functions

Apply differentiation methods in solving problems