Science:Math Exam Resources/Courses/MATH110/December 2010/Question 09 (b)

MATH110 December 2010

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Question 09 (b)
Let $f(x)={\frac {x+1}{x-1}}$. Find all vertical and horizontal asymptotes of $\displaystyle f$, if they exist.

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
Horizontal asymptotes are found by taking the limit of the function to infinity.

Hint 2
Vertical asymptotes consist of all points $x=a$ such that $\lim _{x\to a}f(x)$ diverges to $\pm \infty$. To find values where vertical asymptotes might exist, consider those points where the function is undefined.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

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. To find the horizontal asymptotes, we evaluate the following limit: $\lim _{x\to \pm \infty }{\frac {x+1}{x-1}}$ To evaluate this limit, we need only consider the leading terms - the x in both the numerator and denominator. If we think of factoring out the x from top and bottom and then canceling, we get: $\lim _{x\to \pm \infty }{\frac {x(1+{\frac {1}{x}})}{x(1-{\frac {1}{x}})}}=\lim _{x\to \pm \infty }{\frac {1+{\frac {1}{x}}}{1-{\frac {1}{x}}}}$ The two ${\frac {1}{x}}$ terms go to zero as x goes to plus or minus infinity, so the limit is as follows: $\lim _{x\to \pm \infty }{\frac {1+{\frac {1}{x}}}{1-{\frac {1}{x}}}}={\frac {1}{1}}=1$ So $f(x)$ has a horizontal asymptote, given by the equation $y=1$. Now to find the vertical asymptotes, we need to find all values $x=a$ such that $\lim _{x\to a}f(x)$ diverges to $\pm \infty$. The only candidate for a vertical asymptote for this function is the point where the function is undefined, $x=1$, found in part (a). Indeed, the limit $\lim _{x\to 1}{\frac {x+1}{x-1}}$ goes to infinity. How does this happen? First consider that when the denominator of a fraction is small, the entire fraction is large (see below for an example). In our limit, as the x-values get closer and closer to 1, the number in the denominator is getting smaller and smaller, closer and closer to zero. Using the previous fact about fractions, as the denominator becomes smaller, the entire fraction grows, meaning that the entire function is going to positive or negative infinity. So $f(x)$ has a vertical asymptote at $x=1$. Example: The denominator of the fraction ${\frac {1}{.01}}$ is small, but when it is simplified as follows: ${\frac {1}{.01}}={\frac {1}{\frac {1}{100}}}=100$ we can see that entire function is large - in fact it's a large integer, not even a fraction anymore.

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