Q.Is the function defined by $f(x) = x^2 - \sin x + 5$ continuous at $x = \pi$?

The key idea is continuity at a point: a function $f$ is continuous at $x = a$ if $\lim_{x \to a} f(x) = f(a)$.

1. Compute $f(\pi)$:

   $f(\pi) = \pi^2 - \sin \pi + 5 = \pi^2 - 0 + 5 = \pi^2 + 5$.

2. Find $\lim_{x \to \pi} f(x)$. Since $x^2$, $\sin x$, and the constant $5$ are all continuous everywhere, their combination is continuous. So the limit equals the function value: …

The idea of continuity at a point is simple: as you walk along the graph and approach that point from either side, the function’s output should settle down to exactly the value it has at that point — no jumps, no holes, no wild oscillations. For a function built from familiar pieces like polynomials and trigonometric functions, the usual path is to check three things: the function is defined at the point, the limit exists there, and the two match.

Here, $f(x) = x^2 - \sin x + 5$ is a combination of a polynomial ($x^2 + 5$) and a sine term ($-\sin x$). Both are continuous everywhere on $\mathbb{R}$, so their sum is also continuous everywhere. That already tells us the answer, but let’s verify it step by step — exam questions often expect you to show the reasoning explicitly.

1. Check that $f(\pi)$ is defined. Plug $x = \pi$ directly into the formula:

Since $\sin \pi = 0$, this simplifies to $\pi^2 + 5$. The function is clearly defined — no division by zero or other trouble.

2. Find the limit as $x \to \pi$. Because $x^2$, $\sin x$, and the constant $5$ are all continuous at $\pi$, we can evaluate the limit by direct substitution:

No need for left- and right-hand limits separately — the function is well-behaved enough that the two-sided limit exists and equals this value. …

Method: Checking Continuity by Direct Substitution (Algebra of Continuous Functions)

Use this for any function built purely from standard continuous building blocks — polynomials, $\sin x$, $\cos x$, exponentials, etc., combined by addition, subtraction, multiplication — evaluated at a single ordinary point (not a piecewise boundary).

Steps

Step 1: Confirm the function has no piecewise break or undefined operation at the given point

If $f$ is a sum/difference/product of functions that are each individually continuous everywhere (polynomials, $\sin x$, $\cos x$), and there's no division that could vanish, there is no reason to expect a discontinuity.

Step 2: Invoke the algebra-of-continuous-functions theorem …

Common Mistakes

Mistake 1: Assuming a trig function might break continuity just because it evaluates to zero at that point

Why it's wrong: $\sin \pi = 0$ can look like a "special" or risky value, tempting a student to suspect a hidden discontinuity, but zero is a perfectly ordinary output — $\sin x$ has no breaks or restricted domain anywhere on $\mathbb{R}$. Correct approach: recognize that only situations like division by a vanishing denominator threaten continuity; a trig function simply equaling zero is not one of them.

Mistake 2: Re-deriving the limit using left- and right-hand limits separately when it isn't necessary …