I cannot directly evaluate, since that would lead to division by zero. However, I can logically analyze this limit. The numerator is fixed and the denominator is getting very small. Division by smaller and smaller numbers leads to larger and larger numbers. I conclude the limit must be infinite.The denominator can be negative or positive, so we write
The numerator and denominator both approach zero. This is an indeterminate form of type \(\frac{0}{0}\text{.}\) I need to do some manipulations to deal with it. Both terms are quadratic, so I will try to factor them to simplify the limit.
Both the numerator and denominator approach zero, so this is an indeterminate form of type \(\frac{0}{0}\text{.}\) I can’t do anything to the numerator, but I can factor the denominator to try to simplify the limit. The denominator factors using the special pattern for sums of cubes.
Both numerator and denominator approach zero, so this is an indeterminate form of type \(\frac{0}{0}\text{.}\) I can factor \(x^2\) out of both numerator and denominator to simplify the limit. After that factoring, I can direclty evaluate.
Both numerator and denominator approach 0, so this is a limit of indeterminate form \(\frac{0}{0}\text{.}\) Both terms are quadratics, so I will try to factor the quadratic to simplify the limit.
This limit includes \(x\) and \(\sin x\text{.}\) I will use the known limit for the ratio of these two functions. I need to manipulate this limit to isolate the known limit, using the rules for limits and multiplication.
Then the first part of this is the known limit, which evaluates to 1. The second part can be directly evaluated to 0. The product of 1 and 0 is 0, so the limit is 0.
Activity4.4.8.
Calculate the limit.
\begin{equation*}
\lim_{x \rightarrow 0} x \cot x
\end{equation*}
Solution.
This involves trigonometry and \(x\text{,}\) so I will try to isolate the ration of \(x\) and \(\sin x\text{.}\) To do that I have to write cotagents in terms of sine and cosine.
\begin{equation*}
\lim_{x \rightarrow 0} x \cot x = \lim_{x \rightarrow 0} x
\frac{\cos x}{\sin x}
\end{equation*}
Then I manipulate the limit to isolate the known ratio of \(x\) amd \(sin x\text{.}\)
To find vertical asymptotes of a rational function, I look for the points on the edge of the domain and take the limits approaching those points. Here, domain is all real number except \(x=3\) and \(x=1\text{.}\) Those two numbers are the roots of the denominator. To determine whether there are vertical asymptotes at the undefined points, I need to take the limits approaching them. I have two limits to consider.
Both of these limits are limits where the numerator is fixed and the denominator approaches 0. In these cases, dividing by very small numbers produced very large numbers, so the limits are both \(\pm \infty\text{.}\) Since the limits are infinite, I conclucde that the lines \(x=1\) and \(x=3\) are both vertical asymptotes.
Activity4.4.10.
Find the vertical asymptotes of the the function
\begin{equation*}
f(x) = \frac{x-3}{x^2 + x - 12} \text{.}
\end{equation*}
Solution.
To find vertical asymptotes of a rational function, I look for the points on the edge of the domain and take the limits approaching those points. Here, domain is all real number except \(x=3\) and \(x=-4\text{.}\) Those two numbers are the roots of the denominator. To determine whether there are vertical asymptotes at the undefined points, I need to take the limits approaching them. I have two limits to consider.
In the first limit, the \((x-3)\) terms cancel out and the limit can be directly evaluted to \(\frac{1}{7}\text{.}\) Since this limit if not infinite, this is not a vertical asymptote. In the second limit, the numerator is fixed and the denominator approaches 0. In this case, dividing by very small numbers produced very large numbers, so the limit is \(\pm \infty\text{.}\) Since the limit is infinite, I conclucde that the line \(x=-4\) is is a vertical asymptote.
Unlike the previous questions, this is more about familiarity with common functions and shifts in the plane. I know that the logarithm has an asymptote at \(x=0\text{.}\) Replacing \(x\) by \((x-4)\) is a shift of 4 units in the positive direction, so this function has a vertical asymptote at \(x=4\text{.}\)
This is also about familiarity with known functions. I know the tangent function has asymptotes at every odd multiple of \(\frac{\pi}{2}\text{.}\) In this function, I multiply by 3 inside the tangent function. This multiplication moves the asymptotes closer together. There are now asymptotes at every odd multiple of \(\frac{\pi}{6}\text{.}\)
Subsection4.4.3Continuity
Activity4.4.13.
Find a value for \(a\) (if there is any) which makes this piecewise function continuous.
\begin{equation*}
f(x) = \left\{ \begin{matrix}
x+3 \amp x > 0 \\
-2x + a \amp x \leq 0
\end {matrix} \right.
\end{equation*}
Solution.
To determine continuity, I need to take the limit from the left and the limit from the right approaching the cross over point. Here are the two limits, both of which can be evaluated directly.
To determine continuity, I need to take the limit from the left and the limit from the right approaching the cross over point. Here are the two limits, both of which can be evaluated directly.
To determine continuity, I need to take the limit from the left and the limit from the right approaching the cross over point. Here are the two limits, both of which can be evaluated directly.
This equation is solved when \(a = \frac{-29}{10}\text{.}\)
Activity4.4.16.
Find a value for \(a\) (if there is any) which makes this piecewise function continuous.
\begin{equation*}
f(x) = \left\{ \begin{matrix}
\frac{1}{x} \amp x > 0 \\
x^2 + ax + a \amp x \leq 0
\end {matrix} \right.
\end{equation*}
Solution.
To determine continuity, I need to take the limit from the left and the limit from the right approaching the cross over point. Here are the two limits. The second can be evaluated directly, and the first diverges by logical analysis of the pieces.
I cannot equate these limits, since \(\infty\) is not a number. There is no value of \(a\) which makes this piecewise function continuous.
Subsection4.4.4Limits and Models
Activity4.4.17.
Consider a population model.
\begin{equation*}
f(t) = \left\{ \begin{matrix}
\frac{1}{16}e^{\frac{1}{10}t} \amp t \in [0,10] \\
\frac{39}{12}e^{\frac{1}{7}t} \amp t \in [10,20]
\end{matrix} \right.
\end{equation*}
Is this model continuous? If not, what happens at \(t =
10\text{?}\)
Solution.
I take the limits from the left and from the right to determine continuity. Both limits can be evaluated directly: the limit from the left is \(\frac{e}{16}\) from the right is \(\frac{39}{12}e^{\frac{10}{7}}\text{.}\) The approximate values of these numbers is, respectively, \(0.1699\) and \(13.5614\text{.}\) These are not the same, so the function is not continuous. At \(t=10\) the population has a sudden and significant jump; through some process, a large population has suddenly been added to the existing population.
Activity4.4.18.
Consider a model of temperature in a chemical reaction, as a function of time.
Even though the function is undefined, the limit is reasonable and the temperature simply approaches 24. This tells me that even though the domain is undefined past this point, the model has reasonable and predicatable behaviour.
Subsection4.4.5Conceptual Review Questions
What is a limit?
Why don’t I write division by zero when doing limits?
When do I stop writing \(\lim_{x \rightarrow a}\) in the process of a limit calculation and why?
What is a vertical asymptote? How does it relate to a limit?
The numerator is fixed and the denominator approaches zero. A fixed number divided by a smaller and smaller number produces a larger and larger number, so this limit diverges to infinity. The denominator factors as \((x-4)^2\text{;}\) it is a square, so it is always positive. Therefore, the limit diverges specifically to positive infinity.
First, I can take the limit inside the square root. (This is an application of the limit laws with exponents, since the square root is the same as the exponent \(\frac{1}{2}\text{.}\))
Both numerator and denominator approach zero, so this is an indeterminate form of type \(\frac{0}{0}\text{.}\) I can factor the denominator and cancel off a term to simplify. Then I can directly evaluate.
I can’t even evaluate the pieces here, since there is division by zero in both of the fractions in the numerator. However, I can try to use common denominator and simplification of the nested fraction to put this is a more reasonable form.
Then I can analyze the pieces. The numerator approaches \(-1\text{.}\) The denominator approach \(0\text{.}\) Something close to \(-1\) divided by smaller and smaller numbers produces larger and larger numbers. Therefore, I conclude that the limit diverges to infintiy. It could be positive or negative, since the denominator involves both very small positive and negative values approaching zero.
The numerator approaches \(1\) and the denominator approach \(0\text{.}\) Dividing by smaller and smaller numbers produces larger and larger numbers, so this limit must diverge to inifnity. The numerator is postiive and the denominator, since it is a square, is also positive, so I know, specifically, that this diverges to positive infinity.
This involves trigonometry and \(x\text{,}\) so I will try to use the know limit that relates \(x\) and \(\sin x\text{.}\) To do that, I’ll write tangents as the ration of sine over cosine.
Now I have isolated three limits that I know how to calculate. The limit with the cosine can be directly evaluated. The other two are both the known limit that evalutes to \(1\text{.}\)
