The geometric significance of this result is that the line is a horizontal asymptote for the graph of the function .
Find limits at infinity.
1 End Behavior
In this section we consider limits as approaches either or . There is a connection to the value of these limits and horizontal asymptotes.
(2) If the limit then the line is a horizontal asymptote for the graph of on the left end.
Hence This is evident from the graph of shown below.
The graph of the function also provides evidence for this conclusion.
The geometric significance of this result is that the line is a horizontal asymptote for the graph of the function .
An important generalization of the last example and problem, which follows from a similar analysis is for any constant, , and any positive exponent, .
We will exploit this important fact in the next two examples.
First note that as , a polynomial, according to it leading term. In this example, since the lead terms and both go to as , our limit has the indeterminate form . To resolve this issue, we will factor out of both the numerator and the denominator the highest power of seen in the denominator. So, in this example, we will factor from both. In the numerator, and in the denominator,
With these factorizations, our limit becomes
\begin{align*} \lim _{x \to \infty } \frac{3x^2 + 5x + 2}{2x^2 -x- 4} &= \lim _{x \to \infty } \frac{x^2\left (3 + \frac{5}{x} + \frac{2}{x^2}\right )}{x^2\left (2- \frac{1}{x} - \frac{4}{x^2}\right )} \\ \\ &=\lim _{x \to \infty } \frac{3 + \frac{5}{x} + \frac{2}{x^2}}{2- \frac{1}{x} - \frac{4}{x^2}} \\ \\ &=\frac{3 + 0 + 0}{2- 0 - 0} \\ \\ &= \frac 32. \end{align*}
The result of this limit means that the line is a horizontal asymptote for the graph of (on the right end).
The geometric significance of this result is that the line is a horizontal asymptote for the graph of the function (on the left end).
The geometric significance of this result is that the line is a horizontal asymptote for the graph of the function (on the right end).
First note that as , a polynomial, according to it leading term. In this example, the lead terms and go to and respectively, as . Hence, our limit has the indeterminate form . To resolve this indeterminate form, we will factor out the highest power of in the denominator, namely , from both the numerator and the denominator. In the numerator we get, and in the denominator we get,
Factoring the and canceling, the limit can be resolved as follows:
\begin{align*} \lim _{x \to -\infty }\frac{4x^2 - 3x - 6}{x^3 +8x^2 -3x + 7} &= \lim _{x \to -\infty } \frac{x^3\left (\frac{4}{x}-\frac{3}{x^2} -\frac{6}{x^3}\right )}{x^3\left (1 + \frac{8}{x}-\frac{3}{x^2}+\frac{7}{x^3}\right )} \\ \\ &=\lim _{x \to -\infty } \frac{\frac{4}{x}-\frac{3}{x^2} -\frac{6}{x^3}}{1 + \frac{8}{x}-\frac{3}{x^2}+\frac{7}{x^3}} \\ \\ &=\frac{0 - 0 - 0}{1 +0- 0 + 0} \\ \\ &= \frac 01 \\ &= 0. \end{align*}
The result of this limit means that the line is a horizontal asymptote for the graph of (on the left end).
The result of this limit means that the line is a horizontal asymptote for the graph of (on the left end).
The result of this limit means that the line is a horizontal asymptote for the graph of (on the right end).
Compute the limit:
Suppose is a large negative number. Since is negative, and so using the definition of negative exponents: .
Now note that as and so
Thus, hence
This means that the line is a horizontal asymptote of the graph of , as shown in the graph below.
The value of the limit is (type infinity for , -infinity for or DNE)
The result of this limit means that the line is a horizontal asymptote for the graph of (on the right end).
As increases, the sine function oscillates between and without approaching any particular value. Hence due to oscillation. This is illustrated in the graph of shown below.