Science:Math Exam Resources/Courses/MATH110/April 2018/Question 07 (b)

MATH110 April 2018

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Question 07 (b)
Let $f(x)={\frac {e^{x}}{x}}$. (b) Find all horizontal asymptotes, if they exist. Recall that a function may have different asymptotic behaviours as $x\to +\infty$ and $x\to -\infty$. Make sure you discuss both cases.

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Recall that $y=a$ is a horizontal asymptote of a function $f$ if and only if $\lim _{x\to +\infty }f(x)=a$ or $\lim _{x\to -\infty }f(x)=a$

Hint 2
Use l'Hopital's rule.

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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. First consider the limit of $f$ as $x$ approaches to the positive infinity. Since $\lim _{x\to +\infty }e^{x}=+\infty$ and $\lim _{x\to +\infty }x=+\infty$, the limit of $f$ as $x\to +\infty$ has an ${\frac {\infty }{\infty }}$ indeterminate form. Therefore, we apply l'Hopital's rule to get $$\lim _{x\to +\infty }f(x)=\lim _{x\to +\infty }{\frac {e^{x}}{x}}=\lim _{x\to +\infty }{\frac {(e^{x})'}{(x)'}}=\lim _{x\to +\infty }{\frac {e^{x}}{1}}=+\infty .$$ This means that we don't get a horizontal asymptote of $f$ from the limit $x\to +\infty$ of $f$. On the other hand, when $x$ approaches to the negative infinity, we have $$\lim _{x\to -\infty }e^{x}=0,\quad \lim _{x\to -\infty }{\frac {1}{x}}=0.$$ Using the product rule of limits, we get $$\lim _{x\to -\infty }f(x)=\lim _{x\to -\infty }e^{x}\cdot {\frac {1}{x}}=\lim _{x\to -\infty }e^{x}\cdot \lim _{x\to -\infty }{\frac {1}{x}}=0\cdot 0=0.$$ Therefore, $y=0$ is the only horizontal asymptote of the function $f$. Answer: $\color {blue}y=0$