3397e5…: “Up a fair line”

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Several students are asked to determine if $\Lim {x,y}{3,1} \frac {x-2y}{3x+4y}$ exists and provide the responses below.

Student $1$ explains:

We will analyze the function along different paths.

If $x=0$ , then $F(x,y) = F(0,y) = \frac {-2y}{4y} = -\frac {1}{2}$ .
If $y=0$ , then $F(x,y) = F(x,0) = \frac {x}{3x} = \frac {1}{3}$ .

Since the function approaches different values along different paths, the limit does not exist.

Student $2$ explains:

We will analyze the function along paths of the form $y=mx$ . Along $y=mx$ , we have

$F(x,y)=F(x,mx) = \frac {x-2mx}{3x-4mx} = \frac {1-2m}{3-4m}.$

Since the value the function approaches depends on $m$ , the limit does not exist.

Which student, if either, is correct?

Student $1$ is correct, but Student $2$ is wrong. Student $1$ is wrong, but Student $2$ is correct. Both students are correct. Both students are wrong.

Let’s explore what went wrong. In order to determine if the given limit exists, we need to find a path along which $\vector {x,y} \to \vector {\answer {3},\answer {1}}$ .

Let’s consider Student $1$ first.
In their argument, student $1$ attempted to use a path along which $x=0$ . Can $\vector {x,y}$ approach $\vector {3,1}$ along this path? YesNo
Student $1$ also attempted to use a path along which $y=0$ . Can $\vector {x,y}$ approach $\vector {3,1}$ along this path? YesNo
Now let’s look at Student $2$ ’s argument. In the argument, Student $2$ tried to analyze the function with the most general straight line path. However, note that $y=mx$ is the most general straight line path through $(0,0)$ ; there’s only one value for $m$ for which $y=mx$ passes through $(3,1)$ , and that is $m= \answer {\frac {1}{3}}$ .
The most general straight line path through $(3,1)$ is found using
$y-y_0 = m(x-x_0),$ which in this case is $y-\answer {1} = m\left (x-\answer {3}\right )$ .

That being said, how should we attempt to determine if $\Lim {x,y}{3,1} \frac {x-2y}{3x+4y}$ exists? If we simply try direct substitution, we find that the limit actually does exist and

$\Lim {x,y}{3,1} \frac {x-2y}{3x+4y} = \answer {\frac {1}{13}}.$

There are many techniques that can be used to determine whether limits exist or not, but remember to take a step back, take a deep breath, and think about limits before you start trying to evaluate them.