Problem-Solving Strategy for Continuity at a Point

TL;DR

To check if a function is continuous at a specific point, you need to verify three conditions: the function must be defined at that point, the limit must exist there, and the limit's value must equal the function's value. This systematic approach helps you break down complex problems into manageable steps. Mastering these steps ensures you can confidently determine continuity for any function at a given point.

1. The Mental Model

Think of continuity at a point like drawing a graph without lifting your pen. If you have to lift your pen, there's a break or a hole, meaning it's not continuous. Mathematically, this "smoothness" requires three specific conditions to be met right at that point.

2. The Core Material

Checking for continuity at a specific point, say $x=c$, involves a three-part test. All three conditions must be true for the function $f(x)$ to be continuous at $x=c$. If even one condition fails, the function is not continuous at that point.

The Three Conditions for Continuity at a Point ($x=c$)

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1. $f(c)$ must be defined. This means that when you plug $c$ into the function, you get a real number as an output. There's no hole or undefined value at $x=c$.

2. $\lim_{x \to c} f(x)$ must exist. This means that as $x$ gets closer and closer to $c$ from both the left and the right sides, the function's output $f(x)$ approaches the same finite value. In other words, the left-hand limit must equal the right-hand limit: $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x)$.

3. $\lim_{x \to c} f(x) = f(c)$. This is the final and crucial check. It ensures that the value the function approaches as $x$ gets close to $c$ is exactly the actual value of the function at $x=c$. This condition essentially plugs any "holes" and ensures there's no "jump" or "gap" where the limit exists but doesn't match the function's actual value.

Here's how you can visualize this strategy:

graph TD
    A["Start: Check continuity of f(x) at x=c"] --> B{"Is f(c) defined?"};
    B -- "No" --> D["f(x) is NOT continuous at x=c"];
    B -- "Yes" --> E{"Does lim(x->c) f(x) exist?"};
    E -- "No" --> D;
    E -- "Yes" --> F{"Is lim(x->c) f(x) == f(c)?"};
    F -- "No" --> D;
    F -- "Yes" --> G["f(x) IS continuous at x=c"];

How to Evaluate Limits (Condition 2)

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Often, the trickiest part is evaluating the limit.

• Direct Substitution: If $f(x)$ is a polynomial, rational function (where the denominator isn't zero at $x=c$), or a common transcendental function (like sin, cos, $e^x$, $\ln x$ for positive $x$), you can often just plug $c$ into the function to find the limit. That's because these functions are continuous everywhere in their domains.
• Factoring/Canceling: If direct substitution gives an indeterminate form like $\frac{0}{0}$, try factoring the numerator and denominator, canceling common factors, and then substituting $c$.
• Conjugates: For expressions involving square roots, multiplying by the conjugate can often simplify the expression and allow for direct substitution.
• Piecewise Functions: If your function is defined in different pieces for different intervals, you must check the left-hand limit and the right-hand limit separately at the point where the definition changes. If they match, the limit exists.

3. Worked Example

Let's check the continuity of the function $f(x)$ at $x=1$:
$f(x) = \begin{cases} x^2 + 1 & \text{if } x < 1 \\ 3 & \text{if } x = 1 \\ 2x & \text{if } x > 1 \end{cases}$

We'll go through the three conditions:

Condition 1: Is $f(c)$ defined?
Here, $c=1$. We look at the part of the function definition where $x=1$.
$f(1) = 3$.
Yes, $f(1)$ is defined and equals 3.

Condition 2: Does $\lim_{x \to c} f(x)$ exist?
We need to check the left-hand limit and the right-hand limit as $x$ approaches 1.

• Left-hand limit: As $x \to 1^-$ (meaning $x$ is approaching 1 from values less than 1), we use the top piece of the function: $x^2 + 1$.
  $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^2 + 1) = (1)^2 + 1 = 1 + 1 = 2$.

• Right-hand limit: As $x \to 1^+$ (meaning $x$ is approaching 1 from values greater than 1), we use the bottom piece of the function: $2x$.
  $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2x) = 2(1) = 2$.

Since the left-hand limit (2) equals the right-hand limit (2), the limit exists and $\lim_{x \to 1} f(x) = 2$.

Condition 3: Is $\lim_{x \to c} f(x) = f(c)$?
From Condition 1, we found $f(1) = 3$.
From Condition 2, we found $\lim_{x \to 1} f(x) = 2$.

Is $2 = 3$? No, they are not equal.

Conclusion: Since Condition 3 is not met ($2 \neq 3$), the function $f(x)$ is not continuous at $x=1$.

4. Key Takeaways

• Always check the three conditions in order: function defined, limit exists, limit equals function value.
• For piecewise functions, carefully identify which rule applies to $f(c)$ and for the left/right limits.
• The existence of a limit requires the left-hand and right-hand limits to be equal.
• If direct substitution results in $\frac{0}{0}$, try algebraic manipulation (factoring, conjugates) to simplify before re-evaluating the limit.
• Continuity at a point means the graph has no holes, jumps, or breaks at that specific point.

Common Mistakes to Avoid:
- Forgetting to check all three conditions; missing just one leads to an incorrect conclusion.
- Incorrectly evaluating left-hand and right-hand limits, especially for piecewise functions.
- Assuming a function is continuous just because $f(c)$ is defined or just because the limit exists.
- Mixing up the value of $f(c)$ with the value of $\lim_{x \to c} f(x)$ in Condition 3.

5. Now Try It

For the function $g(x) = \begin{cases} 3x - 2 & \text{if } x \le 2 \\ x^2 & \text{if } x > 2 \end{cases}$, determine if it's continuous at $x=2$. Follow the three-step strategy exactly. What values do you get for each condition, and what's your final conclusion? What success looks like: You should clearly state the result for $g(2)$, $\lim_{x \to 2^-} g(x)$, $\lim_{x \to 2^+} g(x)$, $\lim_{x \to 2} g(x)$, and then compare $\lim_{x \to 2} g(x)$ with $g(2)$ to draw your final conclusion.