Removing Discontinuities
Introduced| English |
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| piecewise function/ˈpiːswaɪz ˈfʌŋkʃn/ |
Patching a hole
- A removable discontinuity is a hole where the limit exists but the value is wrong or missing.
- Since the limit is a perfectly good number, you can redefine the function at that one point to fill the hole.
- Set the point's value equal to the limit — the curve now flows through smoothly.
- This is repairing continuity: possible only for the removable type.
The curve behind the hole
y = x + 3
The cancelled function $y=x+3$ is a straight line — filling the hole just plots the point the line already passes through.
Solving for a mystery constant
- Often a piecewise function 分段函数 has an unknown constant $k$, and you must pick $k$ to make it continuous.
- The rule is condition (3): make the limit equal the value at the seam.
- Set the two pieces equal at the join point and solve for $k$.
- One equation, one unknown — clean algebra.
To make a piecewise function continuous at the seam, set the limit equal to the function ____.
That is continuity condition (3): limit $=$ value.
A worked pattern
- Suppose $f(x)=\begin{cases}x+k,&x<2\\3x-1,&x\ge2\end{cases}$ is to be continuous at $x=2$.
- Left limit: $2+k$. Right value: $3(2)-1=5$.
- Continuity needs $2+k=5$, so $k=3$.
- With $k=3$ the two pieces meet exactly — no jump, a continuous function.
For $f(x)=\begin{cases}x+k,&x<2\\3x-1,&x\ge2\end{cases}$ continuous at $x=2$, find $k$.
Set pieces equal at $2$: $2+k=5\Rightarrow k=3$.
Know what cannot be fixed
- Only removable holes are repairable — the limit must already exist.
- A jump (unequal one-sided limits) has a gap of real width: no single redefinition closes it.
- An infinite discontinuity blows up: no value tames an asymptote.
- So first classify; if the two-sided limit is missing, the discontinuity is permanent.
Which discontinuity can be removed by redefining a single point?
Only the removable type has a two-sided limit to match; jumps and blow-ups cannot be fixed.
Redefining $f(c)$ can create a two-sided limit that did not exist before.
A single point does not affect the limit; if the sides disagree, redefining changes nothing.
Select all discontinuities that cannot be removed.
Anything without a two-sided limit is permanent; only the removable hole can be filled.
You can only "remove" a discontinuity where the two-sided limit already exists. If the left and right limits differ, setting $f(c)$ to any number still leaves the sides disagreeing — the function stays discontinuous. Redefining a point never creates a limit that wasn't there.
To remove the hole of $\dfrac{x^2-25}{x-5}$ at $x=5$, define $f(5)$ equal to the limit. What value?
Limit $=\lim_{x\to5}(x+5)=10$, so $f(5)=10$.
Find $c$ so that $f(x)=\begin{cases}\dfrac{x^2-9}{x-3},&x\neq3\\ c,&x=3\end{cases}$ is continuous at $x=3$.
- The limit: $\displaystyle\lim_{x\to3}\dfrac{x^2-9}{x-3}=\lim_{x\to3}(x+3)=6$.
- Continuity requires $c=\lim_{x\to3}f(x)=6$.
- Setting $c=6$ fills the hole; $f$ is now continuous at $3$.
A removable discontinuity is repaired by redefining $f(c)$ to equal the (existing) limit. For a piecewise function with an unknown constant, set the pieces equal at the join — limit $=$ value — and solve. Jump and infinite discontinuities have no two-sided limit, so they cannot be removed.