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

MATH100 A December 2023

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

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[show] 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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[show] 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 $x\rightarrow \infty$.