Creating a function based on limits

Question

I want to find a function which satisfies certain following limits.

The question is: Find a function which satisfies

$$ \lim_{x\to5} f(x)=3, \text{ and } f(5) \text{ does not exist} $$

I would think that because it says $f(5)$ doesn't exist, there must be a fraction with $(x-5)$ on the bottom. I would think $f(x) = \frac{15}{x-5}$ but that tends to infinity as $x\to5$

Answer 1

Take the following function:

$f:\Bbb R \setminus \{5\} \to \Bbb R$ given by $f(x) =3$

Answer 2

There are, generally, three types of discontinuities: removable, jump, and asymptotic. If you have a jump or asymptotic discontinuity, a finite two-sided limit won't exist.

If you have a jump discontinuity, your left-handed and right-handed limits aren't equal, so the limit doesn't exist.

If you have an asymptotic discontinuity, your left and right-handed limits are each $\pm \infty$, so even if a two-sided limit exists, it's $\pm \infty$, not a finite number. The solution you tried first had this problem - it was asymptotic, and tended to $\infty$ from the right and $-\infty$ from the left.

The simplest way to have a two-sided limit at a point where a function isn't defined is with a removable discontinuity.

As Thomas Andrews noted, a very simple solution is $\frac{3(x-5)}{x-5}$. That fraction is equivalent to $3$ everywhere except at $x=5$, where it has a hole.

You could, in fact, take any function $g$ such that $g$ is continuous at $x = 5$ and $g(5) = 3$, then let $f(x) = g(x)*\frac{x-5}{x-5}$. Similarly, you could define $f$ piecewise, where it's equal to $g$ everywhere except $x = 5$, and undefined at $x = 5$. That's the solution that jjagmath is getting at.

Thomas's solution uses $g(x) = 3$, but you could use, for example, $g(x) = x-2$, or $g(x) = \sqrt{x+4}$, or anything else that satisfies those above conditions.

Answer 3

$$f(x)=\frac{\sin(3(x-5))}{x-5}$$ or $$f(x)=\frac{e^{3(x-5)}-1}{x-5}$$ or $$f(x)=\frac{\ln(3x-14)}{x-5}$$ satifies the wanted conditions.