How do you find horizontal asymptotes using limits?
Answer
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Hint: There are two types of asymptotes namely horizontal asymptotes and vertical asymptotes. We explain the horizontal asymptotes with an example so that we can understand the concept clearly. To find horizontal asymptotes, we simply evaluate the limit of the function as it approaches infinity, and again as it approaches negative infinity.
Complete step-by-step answer:
Horizontal asymptotes:
A function f(x) will have a horizontal asymptote \[y = L\] if either \[\mathop {\lim }\limits_{x \to \infty } f(x) = L\] or \[\mathop {\lim }\limits_{x \to - \infty } f(x) = L\] . That is we need to find the limit value as ‘x’ tends to infinity and negative infinity.
Let’s take an example, \[f(x) = \dfrac{{3x + 7}}{{2x - 5}}\] .
Now we first find the limit value as ‘x’ tends to infinity.
That is \[\mathop {\lim }\limits_{x \to \infty } f(x) = \mathop {\lim }\limits_{x \to \infty } \dfrac{{3x + 7}}{{2x - 5}}\]
Taking ‘x’ common in the numerator and denominator we get
\[ = \mathop {\lim }\limits_{x \to \infty } \dfrac{{x\left( {3 + \dfrac{7}{x}} \right)}}{{x\left( {2 - \dfrac{5}{x}} \right)}}\]
\[ = \mathop {\lim }\limits_{x \to \infty } \dfrac{{\left( {3 + \dfrac{7}{x}} \right)}}{{\left( {2 - \dfrac{5}{x}} \right)}}\]
Applying the limit. As \[x \to \infty \] we will have \[\dfrac{7}{x} \to 0\] and \[\dfrac{5}{x} \to 0\] , we get,
\[ = \dfrac{3}{2}\] .
That is \[\mathop {\lim }\limits_{x \to \infty } f(x) = \dfrac{3}{2}\] .
Similarly find the limit value as ‘x’ tends to negative infinity,
That is \[\mathop {\lim }\limits_{x \to - \infty } f(x) = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{3x + 7}}{{2x - 5}}\]
Taking ‘x’ common in the numerator and denominator we get
\[ = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{x\left( {3 + \dfrac{7}{x}} \right)}}{{x\left( {2 - \dfrac{5}{x}} \right)}}\]
\[ = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{\left( {3 + \dfrac{7}{x}} \right)}}{{\left( {2 - \dfrac{5}{x}} \right)}}\]
Applying the limit. As \[x \to - \infty \] we will have \[\dfrac{7}{x} \to 0\] and \[\dfrac{5}{x} \to 0\] , we get,
\[ = \dfrac{3}{2}\] .
That is \[\mathop {\lim }\limits_{x \to - \infty } f(x) = \dfrac{3}{2}\] .
Therefore, the function has horizontal asymptote at \[y = \dfrac{3}{2}\] .
So, the correct answer is “ \[y = \dfrac{3}{2}\] ”.
Note: Since the two limits were identical, this function has a single horizontal asymptote. Since asymptotes are lines, they are described by equations, not just by numbers. A function can have at most two horizontal asymptotes, one in each direction. Vertical asymptote is when ‘x’ approaches any constant value ‘c’, parallel to the y-axis. Horizontal asymptotes are flat lines parallel to the x-axis.
Complete step-by-step answer:
Horizontal asymptotes:
A function f(x) will have a horizontal asymptote \[y = L\] if either \[\mathop {\lim }\limits_{x \to \infty } f(x) = L\] or \[\mathop {\lim }\limits_{x \to - \infty } f(x) = L\] . That is we need to find the limit value as ‘x’ tends to infinity and negative infinity.
Let’s take an example, \[f(x) = \dfrac{{3x + 7}}{{2x - 5}}\] .
Now we first find the limit value as ‘x’ tends to infinity.
That is \[\mathop {\lim }\limits_{x \to \infty } f(x) = \mathop {\lim }\limits_{x \to \infty } \dfrac{{3x + 7}}{{2x - 5}}\]
Taking ‘x’ common in the numerator and denominator we get
\[ = \mathop {\lim }\limits_{x \to \infty } \dfrac{{x\left( {3 + \dfrac{7}{x}} \right)}}{{x\left( {2 - \dfrac{5}{x}} \right)}}\]
\[ = \mathop {\lim }\limits_{x \to \infty } \dfrac{{\left( {3 + \dfrac{7}{x}} \right)}}{{\left( {2 - \dfrac{5}{x}} \right)}}\]
Applying the limit. As \[x \to \infty \] we will have \[\dfrac{7}{x} \to 0\] and \[\dfrac{5}{x} \to 0\] , we get,
\[ = \dfrac{3}{2}\] .
That is \[\mathop {\lim }\limits_{x \to \infty } f(x) = \dfrac{3}{2}\] .
Similarly find the limit value as ‘x’ tends to negative infinity,
That is \[\mathop {\lim }\limits_{x \to - \infty } f(x) = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{3x + 7}}{{2x - 5}}\]
Taking ‘x’ common in the numerator and denominator we get
\[ = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{x\left( {3 + \dfrac{7}{x}} \right)}}{{x\left( {2 - \dfrac{5}{x}} \right)}}\]
\[ = \mathop {\lim }\limits_{x \to - \infty } \dfrac{{\left( {3 + \dfrac{7}{x}} \right)}}{{\left( {2 - \dfrac{5}{x}} \right)}}\]
Applying the limit. As \[x \to - \infty \] we will have \[\dfrac{7}{x} \to 0\] and \[\dfrac{5}{x} \to 0\] , we get,
\[ = \dfrac{3}{2}\] .
That is \[\mathop {\lim }\limits_{x \to - \infty } f(x) = \dfrac{3}{2}\] .
Therefore, the function has horizontal asymptote at \[y = \dfrac{3}{2}\] .
So, the correct answer is “ \[y = \dfrac{3}{2}\] ”.
Note: Since the two limits were identical, this function has a single horizontal asymptote. Since asymptotes are lines, they are described by equations, not just by numbers. A function can have at most two horizontal asymptotes, one in each direction. Vertical asymptote is when ‘x’ approaches any constant value ‘c’, parallel to the y-axis. Horizontal asymptotes are flat lines parallel to the x-axis.
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