Science:Math Exam Resources/Courses/MATH100 A/December 2023/Question 01

MATH100 A December 2023

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Question 01
List the vertical and horizontal asymptotes of the curve given by $y={\frac {x-4}{\sqrt {4x^{2}+3}}}$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint

Recall the definitions. A function $f$ has a vertical asymptote at $x=c$ if $\displaystyle \lim _{x\rightarrow c^{-}}f(x)=\pm \infty$ or $\displaystyle \lim _{x\rightarrow c^{+}}f(x)=\pm \infty$. What does this mean for $f(x)$ if $x$ is near c? Can $f$ be continuous at $c$? A function $f$ has a horizontal asymptote at $-\infty$ or $\infty$ if the limit $\displaystyle \lim _{x\rightarrow -\infty }f(x)$ exists or if the limit $\lim _{x\rightarrow \infty }f(x)$ exists, respectively. Can you try to compute these limits?

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. If a function $f$ is continuous at $x=c$, then the line $\{x=c\}$ cannot be a vertical asymptote. Our function is a ratio of two continuous functions, so it is continuous wherever the denominator does not vanish. Therefore, the only possible asymptotes are the numbers $c$ for which the denominator vanishes, in other words, those $c$ for which ${\sqrt {4c^{2}+3}}=0$. Since $c^{2}\geq 0$, it follows that $4c^{2}+3\geq 3>0$, so the denominator does not vanish for any $c$. Therefore our function has no vertical asymptote. For the horizontal asymptotes, we must compute (or determine inexistence of) the limits (1) as $x\rightarrow -\infty$ and (2) as $x\rightarrow \infty$. The computations of limits (1) and (2) start off the same: ${\begin{aligned}\lim _{x\rightarrow \pm \infty }{\frac {x-4}{\sqrt {4x^{2}+3}}}&=\lim _{x\rightarrow \pm \infty }{\frac {x-4}{\sqrt {4x^{2}(1+{\frac {3}{4x^{2}}})}}}\\&=\lim _{x\rightarrow \pm \infty }{\frac {x(1-{\frac {4}{x}})}{{\sqrt {4x^{2}}}{\sqrt {1+{\frac {3}{4x^{2}}}}}}}\\&=\lim _{x\rightarrow \pm \infty }{\frac {x(1-{\frac {4}{x}})}{2|x|\times {\sqrt {1+{\frac {3}{4x^{2}}}}}}}\\&=\lim _{x\rightarrow \pm \infty }{\frac {x}{2|x|}}\times {\frac {1+0}{\sqrt {1+0}}}\\&=\lim _{x\rightarrow \pm \infty }{\frac {x}{2|x|}}\end{aligned}}$ To compute (1), notice that, if we are taking the limit as $x\rightarrow -\infty$, then we may assume that $x<0$, so $|x|=-x$, and we have $\lim _{x\rightarrow -\infty }{\frac {x-4}{\sqrt {4x^{2}+3}}}=\lim _{x\rightarrow -\infty }{\frac {x}{2|x|}}=\lim _{x\rightarrow -\infty }{\frac {x}{-2x}}={\frac {-1}{2}}.$ To compute (2), we are in the situation where $x>0$, so $|x|=x$ and we have $\lim _{x\rightarrow \infty }{\frac {x-4}{\sqrt {4x^{2}+3}}}=\lim _{x\rightarrow \infty }{\frac {x}{2x}}={\frac {1}{2}}.$ Thus, there are two horizontal asymptotes: one is the line $\{y=-1/2\}$, which the function approaches as $x\rightarrow -\infty$. The other is the line $\{y=1/2\}$, which the function approaches as $x\rightarrow \infty$.