Question

Which of the following functions have a horizontal asymptote:
(a) $f(x)=\dfrac{{{x}^{3}}}{{{x}^{2}}+5}$
(b) $f(x)={{\tan }^{-1}}x$
(c) $f(x)=\sqrt{x+7}$
(d) $f(x)=5{{x}^{5}}+7{{x}^{2}}-1$

Answer 1

Hint: Remember that a function f(x) is said to have a horizontal asymptote if and only if $\underset{x\to \infty }{\mathop{\lim }}\,f(x)=c$ , where c is finite real number. To find the functions which have horizontal asymptotes, check each option one by one until you get the answer.

Complete step by step answer:
The asymptote of a curve is the tangent to the curve at infinity, i.e. the tangent which actually doesn’t meet the curve but the distance between the curve and the tangent is continuously decreasing and is expected to touch the curve at infinity is known as the asymptotes of a given curve.
For an asymptote to the curve f(x) to be horizontal must satisfy the condition that $\underset{x\to \infty }{\mathop{\lim }}\,f(x)=c$ , where c is finite real number. So, to get the answer to the above question, we will have to check the options one by one. Let us start with option (a).
For option (a), $f(x)=\dfrac{{{x}^{3}}}{{{x}^{2}}+5}$ , so if we take limit of both the sides for x tending to infinity, we get
$\underset{x\to \infty }{\mathop{\lim }}\,f(x)=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{{{x}^{3}}}{{{x}^{2}}+5}$
$\Rightarrow \underset{x\to \infty }{\mathop{\lim }}\,f(x)=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{{{x}^{3}}}{\left( 1+\dfrac{5}{{{x}^{2}}} \right){{x}^{2}}}$
$\Rightarrow \underset{x\to \infty }{\mathop{\lim }}\,f(x)=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{x}{\left( 1+\dfrac{5}{{{x}^{2}}} \right)}$
Now, if we use the property that any natural number divided by infinity is 0 and put the limit value of x in the RHS, we get
$\Rightarrow \underset{x\to \infty }{\mathop{\lim }}\,f(x)=\dfrac{\infty }{1+\dfrac{5}{\infty }}=\dfrac{\infty }{1+0}=\infty $
So, the condition is not satisfied, so option (a) is not the correct answer.
Similarly, let us check for option (b) ${{\tan }^{-1}}x$ .
$f(x)={{\tan }^{-1}}x$
Taking limit of both the sides of the equation, we get
$\underset{x\to \infty }{\mathop{\lim }}\,f(x)=\underset{x\to \infty }{\mathop{\lim }}\,{{\tan }^{-1}}x$
As no simplification possible, we will directly put the limit. On doing so, we get
$\underset{x\to \infty }{\mathop{\lim }}\,f(x)={{\tan }^{-1}}\infty $
We know that ${{\tan }^{-1}}\infty =\dfrac{\pi }{2}$ .
$\underset{x\to \infty }{\mathop{\lim }}\,f(x)=\dfrac{\pi }{2}$
As $\dfrac{\pi }{2}$ is a real number, we can say that $f(x)={{\tan }^{-1}}x$ has an horizontal Asymptote.
Now, let us check for option (c).
$f(x)=\sqrt{x+7}$
Taking limit of both the sides of the equation, we get
$\underset{x\to \infty }{\mathop{\lim }}\,f(x)=\underset{x\to \infty }{\mathop{\lim }}\,\sqrt{x+7}$
It is clear that as x tends to infinity, $\sqrt{x+7}$ also becomes infinity. So, $f(x)=\sqrt{x+7}$ has no horizontal asymptotes.
Finally, let us check for option (d).
$f(x)=5{{x}^{5}}+7{{x}^{2}}-1$
Taking limit of both the sides of the equation, we get
$\underset{x\to \infty }{\mathop{\lim }}\,f(x)=\underset{x\to \infty }{\mathop{\lim }}\,5{{x}^{5}}+7{{x}^{2}}-1$
As no simplification available, we will put the limit. On doing so, we get
$\underset{x\to \infty }{\mathop{\lim }}\,f(x)=\infty +\infty -1=\infty $
So, $f(x)=5{{x}^{5}}+7{{x}^{2}}-1$ has no horizontal asymptotes.
Therefore, the answer to the above question is an option (b).

Note:
Remember that if a curve satisfies the condition $\underset{x\to -\infty }{\mathop{\lim }}\,f(x)=c$ will prove that it has a horizontal Asymptote, but the chances of a function not satisfying $\underset{x\to \infty }{\mathop{\lim }}\,f(x)=c$ but satisfying this $\underset{x\to -\infty }{\mathop{\lim }}\,f(x)=c$ is almost nil, so, it is better that you check all the options for either of the one condition, if you don’t get the answer then go for the other.