When two pieces of a function meet at x = 1
Consider the piecewise function $$ f(x,\alpha) = \begin{cases} \alpha k + 1, & \text{if } x < 1, \\ \sin(x\pi) + 2, & \text{if } x > 1, \end{cases} $$ where $\alpha$ and $k$ are constants. Analyze the behavior of $f$ on each piece and determine the condition on $\alpha$ and $k$ under which the two sides can be joined continuously at $x=1$.
Explanation of Method
Since $f$ is defined by two different formulas on either side of $x=1$ and is left undefined exactly at $x=1$, the way to understand its behavior there is to compute the left-hand and right-hand limits separately. The left piece is a constant, so its limit is just that constant. The right piece involves $\sin(x\pi)$, which is continuous, so its limit is found by direct substitution. Comparing the two one-sided limits then tells us exactly which relationship between $\alpha$ and $k$ (if any) would make the point $x=1$ removable, i.e., would let $f$ extend to a continuous function there.
- Evaluate the left-hand limit, which is the constant value of the first piece.
- Evaluate the right-hand limit using continuity of $\sin$.
- Compare the two one-sided limits to find the condition for a continuous extension at $x=1$.
Step 1 · Behavior for $x < 1$
For $x<1$, $f(x,\alpha) = \alpha k + 1$ is a constant, independent of $x$. So the left-hand limit at $x=1$ is simply that constant:
$$ \lim_{x \to 1^-} f(x,\alpha) = \alpha k + 1. $$
Step 2 · Behavior for $x > 1$
For $x > 1$, $f(x,\alpha) = \sin(x\pi) + 2$. Since $\sin(x\pi)$ oscillates between $-1$ and $1$ for all $x$, this piece takes values in:
$$ f(x,\alpha) \in [1,3] \quad \text{for } x > 1. $$
Since $\sin$ is continuous, the right-hand limit at $x=1$ is found by direct substitution:
$$ \lim_{x \to 1^+} f(x,\alpha) = \sin(\pi) + 2 = 0 + 2 = 2. $$
Step 3 · Compare the one-sided limits
The two one-sided limits are:
$$ \lim_{x \to 1^-} f(x,\alpha) = \alpha k + 1, \qquad \lim_{x \to 1^+} f(x,\alpha) = 2. $$
Since $f$ is not defined at $x=1$ itself (only $x<1$ and $x>1$ are covered), the two-sided limit $\lim_{x\to1} f(x,\alpha)$ exists exactly when these two one-sided limits agree:
$$ \alpha k + 1 = 2 \quad\Longleftrightarrow\quad \alpha k = 1. $$
Whenever $\alpha k = 1$, both one-sided limits equal $2$, so $f$ has a removable "gap" at $x=1$ that could be filled in by defining $f(1,\alpha) = 2$ to make it continuous there. For any other values of $\alpha$ and $k$ (i.e., whenever $\alpha k \neq 1$), the two one-sided limits disagree and $f$ has a genuine jump discontinuity at $x=1$ that cannot be repaired by any single choice of $f(1,\alpha)$.
Summary Table
| Side | Expression | Limit at $x=1$ |
|---|---|---|
| Left ($x<1$) | $\alpha k+1$ (constant) | $\alpha k+1$ |
| Right ($x>1$) | $\sin(x\pi)+2$ | $2$ |
Additional Notes
- Because the left piece is a constant and does not depend on $x$ at all, its "limit" is trivial — the entire question of continuity reduces to whether that constant matches the value the right piece limits to.
- The condition $\alpha k = 1$ describes a whole family of pairs $(\alpha,k)$ — for instance $\alpha=1,\,k=1$ or $\alpha=2,\,k=\tfrac12$ — any of which makes the two pieces meet at the common value $2$.
- This differs from a genuinely oscillating limit: even though $\sin(x\pi)+2$ ranges over $[1,3]$ for $x>1$ in general, its behavior specifically as $x\to1^+$ is governed only by the single value $\sin(\pi)+2=2$, since $\sin$ is continuous there.
Result
The one-sided limits at $x=1$, and the condition for them to agree, are: