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New Zealand
Level 8 - NCEA Level 3

Limits at Infinity


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)=ax+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$\lim_{x\rightarrow\infty}\frac{1}{x}=0$limx1x=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 $\lim_{x\rightarrow0}\frac{\sin x}{x}=1$limx0sinxx=1 and $\lim_{x\rightarrow0}\frac{\cos x-1}{x}=0$limx0cosx1x=0.


Example 1

Find the limit, if it exists, of the function $x\sin\frac{1}{x}$xsin1x as $x\rightarrow\infty$x.

As $x$x becomes arbitrarily large, the expression approaches the form $\infty\times0$×0 which is indeterminate. However, $x\sin\frac{1}{x}$xsin1x can be rewritten as $\frac{\sin\frac{1}{x}}{\frac{1}{x}}$sin1x1x.

We know that $\lim_{y\rightarrow0}\frac{\sin y}{y}=1$limy0sinyy=1 and $\lim_{x\rightarrow\infty}\frac{1}{x}=0$limx1x=0. Now, if we put $y=\frac{1}{x}$y=1x, we have $\lim_{x\rightarrow\infty}\frac{\sin\frac{1}{x}}{\frac{1}{x}}=\lim_{y\rightarrow0}\frac{\sin y}{y}=1$limxsin1x1x=limy0sinyy=1.


Example 2

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)=x32x2x3+x21.

We rewrite the function as $\frac{1-\frac{2}{x^2}}{2+\frac{1}{x}-\frac{1}{x^3}}$12x22+1x1x3 by dividing the numerator and denominator by $x^3$x3. Now, because $\lim_{x\rightarrow\infty}\frac{1}{x}=0$limx1x=0 it must be true (by a limit theorem) that $\lim_{x\rightarrow\infty}\frac{1}{x^2}=0$limx1x2=0 and $\lim_{x\rightarrow\infty}\frac{1}{x^3}=0$limx1x3=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.

Example 3

Find the value of $\lim_{x\to\infty}\left(x\sin\left(\frac{4}{x}\right)\right)$limx(xsin(4x)).

Example 4

Find the value of $\lim_{x\to\infty}\left(x-\sqrt{x^2+7}\right)$limx(xx2+7).

Example 5

Use the graph of $y=2-e^{-x}$y=2ex to find the value of $\lim_{x\to\infty}\left(2-e^{-x}\right)$limx(2ex).

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