Q.Check the continuity of the function $f$ given by $f(x) = 2x + 3$ at $x = 1$.

Concept: Continuity At A Point — A function $f$ is continuous at $x = a$ if $\lim_{x \to a} f(x) = f(a)$.

1. Compute $f(1)$:

   $f(1) = 2(1) + 3 = 5$.

2. Compute $\lim_{x \to 1} f(x)$:

   Since $f(x) = 2x + 3$ is a polynomial, the limit is simply the value at $x = 1$:

   $\lim_{x \to 1} (2x + 3) = 2(1) + 3 = 5$.

3. Compare:

   $\lim_{x \to 1} f(x) = 5 = f(1)$.

All three conditions of continuity are satisfied.

The idea of continuity at a point is simple: a function is continuous at $x=a$ if you can draw its graph near that point without lifting your pen. More formally, three things must be true:

1. The function is defined at $a$ — $f(a)$ exists.
2. The limit of $f(x)$ as $x$ approaches $a$ exists.
3. That limit equals the function value: $\lim_{x\to a} f(x) = f(a)$.

If any one of these fails, the function is discontinuous at that point.

For $f(x)=2x+3$, we're dealing with a straight line — the simplest continuous function there is. Let's check each condition at $x=1$.

Step 1: Check if $f(1)$ exists.

Plug $x=1$ into the function:

The function is defined and gives a value of $5$. Condition 1 is satisfied.

Step 2: Check if $\lim_{x\to 1} f(x)$ exists.

Since $f$ is a polynomial (specifically a linear polynomial), its limit as $x$ approaches any real number is simply the function value at that point. But let's verify from both sides to be thorough.

Left-hand limit ($x \to 1^-$):

Right-hand limit ($x \to 1^+$):

Both one-sided limits are equal to $5$, so the two-sided limit exists and is $5$. Condition 2 is satisfied.

Step 3: Check if $\lim_{x\to 1} f(x) = f(1)$.

We have:

They are equal. Condition 3 is satisfied.

All three conditions hold. Therefore, $f$ is continuous at $x=1$.

Method: Checking Continuity at a Point

A function $f$ is continuous at $x=a$ only if all three of the following hold — checking just one or two is not a complete proof.

Steps

Step 1: Confirm $f(a)$ exists

Evaluate the function directly at the point.

Step 2: Confirm $\displaystyle\lim_{x\to a}f(x)$ exists

For most elementary functions this means checking that the left-hand limit and the right-hand limit are equal.

Step 3: Confirm the limit equals the function value

If all three conditions hold, $f$ is continuous at $a$; if any one fails, it is not.

Applying to this problem: for $f(x)=2x+3$ at $x=1$, $f(1)=5$, and both one-sided limits (and hence the two-sided limit) equal $5$ since linear functions have no breaks — all three conditions hold, so $f$ is continuous at $x=1$.

Common Mistakes

Mistake 1: Skipping the one-sided-limit check because the function "looks obviously continuous".

Why it's wrong: even for a simple linear function, a rigorous continuity proof requires demonstrating the limit exists (via matching one-sided limits), not just asserting it from visual intuition — a board exam expects the explicit check. Correct approach: always state the left-hand and right-hand limits separately, even when they're evidently going to match.

Mistake 2: Confusing "continuous" with "differentiable" and answering the wrong question.

Why it's wrong: continuity only requires the limit to equal the function value — no statement about the function's derivative is needed or relevant here. Correct approach: keep the three-condition continuity check entirely separate from any discussion of differentiability.