Determining a parameter so that it's a removable singularity

[greg_rack]

Homework Statement

Determine $a$ so that $f(x)=\frac{ax^2-16}{x^2-6x+8}$ has a removable singularity for $x=2$

Relevant Equations

none

So, for this exercise I'm considering the removable singularity for $x=2$ to cause $f(2)$ to be different from $\lim_{x \to 2}f(x)$.
But as soon as I write everything down, I get stuck here: $\lim_{x \to 2}f(x)\neq\frac{4(a-4)}{0}$
How do I calculate $a$?

 

[anuttarasammyak]

$$\frac{ax^2-16}{(x-4)(x-2)}=\frac{...?}{x-4}$$

 

[PeroK]

So, for this exercise I'm considering the removable singularity for $x=2$ to cause $f(2)$ to be different from $\lim_{x \to 2}f(x)$.

This isn't what you want. You need the limit to exist. Then, you can remove the singuality by defining $f(2)$ to be the value of the limit.

 

[greg_rack]

$$\frac{ax^2-16}{(x-4)(x-2)}=\frac{...?}{x-4}$$

Mmm, I didn't get that... Have you factored the numerator to simplify that $(x-2)$ at the denominator? If yes, why? And how, since there is an $'a'$?

 

[greg_rack]

This isn't what you want. You need the limit to exist. Then, you can remove the singuality by defining $f(2)$ to be the value of the limit.

Wait... I do need a singularity, a so-called "removable" one. If $f(2)=\lim_{x \to 2}f(x)$, the function would be continuous in 2, and that's not what I want. Correct?

EDIT, from Wikipedia: In analysis, a removable singularity of a function is a point at which the function is undefined.

 

[PeroK]

Wait... I do need a singularity, a so-called "removable" one. If $f(2)=\lim_{x \to 2}f(x)$, the function would be continuous in 2, and that's not what I want. Correct?

That is what you want. A removable singularity means the limit exists and allows you to create a continuous function.

Note that whether the $\lim_{x \rightarrow 2} f(x)$ exists does not depend on $f(2)$.

Whatever you do $f(2)$ cannot be defined by evaluating the given expression. That will always have $0$ on the denominator, hence be undefined.

 

[greg_rack]

That is what you want. A removable singularity means the limit exists and allows you to create a continuous function.

Note that whether the $\lim_{x \rightarrow 2} f(x)$ exists does not depend on $f(2)$.

And so how do I establish $a$? Which conditions should I use

 

[PeroK]

And so how do I establish $a$? Which conditions should I use

The limit must exist. The answer is screaming at you from your original post!

 

[anuttarasammyak]

If yes, why? And how, since there is an ′a′?

Because if (x-2) disappears from denominator you do not have to worry about singularity at x=2.
Find a good value of 'a' so that you can do it.

 

[greg_rack]

The limit must exist. The answer is screaming at you from your original post!

God, I'm feeling so dumb!
The limit should therefore exist, independently from what the value of $f(2)$ is, so as @anuttarasammyak said I should find the value of $a$ for which $\frac{ax^2-16}{(x-4)(x-2)}$ exists(?)

 

[PeroK]

The limit should therefore exist, independently from what the value of $f(2)$ is, so as @anuttarasammyak said I should find the value of $a$ for which $\frac{ax^2-16}{(x-4)(x-2)}$ exists(?)

Yes, you need that limit to exist as $x \rightarrow 2$.

 

[greg_rack]

Yes, you need that limit to exist as $x \rightarrow 2$.

Ok then, and what about $f(2)$? My textbook says it should not exist or at least be different from the value of the limit.

 

[PeroK]

Ok then, and what about $f(2)$? My textbook says it should not exist or at least be different from the value of the limit.

$f(2)$ is undefined if all you have is that formula for $f(x)$:
$$f(2) = \frac{4a - 16}{0}$$
Which is undefined, no matter what $a$ is. The function you have been given is not defined at $x =2$ or $x =4$. If we take $a = 0$, then $f(x) = \frac{-16}{(x-2)(x-4)}$.

That has non-removable singularities at $x = 2$ and $x = 4$. This is because the function blows up around $x = 2$ and $x = 4$. Note that it's the behaviour of the function close to these points that is important.

You need to find some $a$ which makes the function well behaved close to $x = 2$.

 

[greg_rack]

You need to find some a which makes the function well behaved close to x=2.

I'm finally getting there! Last easy and merely algebraical question: how do I determine $a$ so that the factor which gives indetermination($(x-2)$) in $$\frac{ax^2-16}{(x-2)(x-4)}$$ can be simplified?

 

[PeroK]

I'm finally getting there! Last easy and merely algebraical question: how do I determine $a$ so that the factor which gives indetermination($(x-2)$) in $$\frac{ax^2-16}{(x-2)(x-4)}$$ can be simplified?

What would you like to get rid of from the denominator? What do you need on the numerator?

 

[greg_rack]

What would you like to get rid of from the denominator? What do you need on the numerator?

I should get rid of $(x-2)$ by having another $(x-2)$ on the numerator, right?