Understanding Continuity of a Function

For a function \({\rm{f}}({\rm{x}})\) to be continuous at a point \({\rm{x}} = {\rm{a}}\), three conditions must be met:

1. The function \({\rm{f}}({\rm{a}})\) must be defined at \({\rm{x}} = {\rm{a}}\).
2. The limit of the function as \({\rm{x}}\) approaches \({\rm{a}}\), i.e., \(\lim_{x \to a} f(x)\), must exist.
3. The value of the function at the point must be equal to the limit of the function at the point, i.e., \({\rm{f}}({\rm{a}}) = \lim_{x \to a} f(x)\).

In this problem, the function is given by \({\rm{f}}\left( {\rm{x}} \right) = \frac{{1 - \sin {\rm{x}} + \cos {\rm{x}}}}{{1 + \sin {\rm{x}} + \cos {\rm{x}}}}\). We are told that the function is not defined at \({\rm{x}} = \pi\). To make the function continuous at \({\rm{x}} = \pi\), we need to define \({\rm{f}}(\pi)\) such that \({\rm{f}}(\pi) = \lim_{x \to \pi} f(x)\).

Calculating the Limit for Continuity

We need to find the limit of the function as \({\rm{x}}\) approaches \({\rm{\pi}}\):

\(\lim_{x \to \pi} \frac{{1 - \sin {\rm{x}} + \cos {\rm{x}}}}{{1 + \sin {\rm{x}} + \cos {\rm{x}}}}\)

Let's substitute \({\rm{x}} = \pi\) into the expression:

• \(\sin \pi = 0\)
• \(\cos \pi = -1\)

Substituting these values, we get:

Numerator: \(1 - \sin \pi + \cos \pi = 1 - 0 + (-1) = 0\)

Denominator: \(1 + \sin \pi + \cos \pi = 1 + 0 + (-1) = 0\)

The limit is of the indeterminate form \(\frac{0}{0}\). We can use L'Hopital's Rule or trigonometric identities to evaluate this limit.

Applying L'Hopital's Rule

L'Hopital's Rule states that if \(\lim_{x \to a} \frac{g(x)}{h(x)}\) is of the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then \(\lim_{x \to a} \frac{g(x)}{h(x)} = \lim_{x \to a} \frac{g'(x)}{h'(x)}\), provided the latter limit exists.

Let \(g(x) = 1 - \sin x + \cos x\) and \(h(x) = 1 + \sin x + \cos x\).

Find the derivatives:

• \(g'(x) = \frac{d}{dx}(1 - \sin x + \cos x) = 0 - \cos x - \sin x = -\cos x - \sin x\)
• \(h'(x) = \frac{d}{dx}(1 + \sin x + \cos x) = 0 + \cos x - \sin x = \cos x - \sin x\)

Now apply L'Hopital's Rule:

\(\lim_{x \to \pi} \frac{-\cos x - \sin x}{\cos x - \sin x}\)

Substitute \({\rm{x}} = \pi\):

\(\frac{-\cos \pi - \sin \pi}{\cos \pi - \sin \pi} = \frac{-(-1) - 0}{-1 - 0} = \frac{1}{-1} = -1\)

So, the limit of the function as \({\rm{x}}\) approaches \({\rm{\pi}}\) is \(-1\).

Determining f(π) for Continuity

For the function \({\rm{f}}({\rm{x}})\) to be continuous at \({\rm{x}} = \pi\), the value of \({\rm{f}}(\pi)\) must be equal to the limit we just calculated.

\({\rm{f}}(\pi) = \lim_{x \to \pi} f(x) = -1\)

Therefore, the value of \({\rm{f}}(\pi)\) must be \(-1\) to make the function continuous at \({\rm{x}} = \pi\).

Summary of the Solution

The value of \({\rm{f}}(\pi)\) required for continuity at \({\rm{x}} = \pi\) is \(-1\).

Revision Table: Key Concepts

Additional Information: Understanding Continuity and Limits

Continuity is a fundamental concept in calculus. Informally, a function is continuous on an interval if you can draw its graph over that interval without lifting your pen. A break or a hole in the graph indicates a point of discontinuity.

There are different types of discontinuity:

• Removable Discontinuity: This occurs when the limit of the function exists at a point, but the function is either not defined at that point or its value at the point is not equal to the limit. In our problem, the discontinuity at \(x = \pi\) is removable because the limit exists (\(-1\)). By defining \(f(\pi) = -1\), we 'remove' the discontinuity.
• Jump Discontinuity: This happens when the left-hand limit and the right-hand limit at a point exist but are not equal.
• Infinite Discontinuity: This occurs when the function approaches positive or negative infinity as \(x\) approaches the point. This usually involves a vertical asymptote.

Limits are essential for defining continuity, derivatives, and integrals. Evaluating limits often involves algebraic simplification, using trigonometric identities, or applying rules like L'Hopital's Rule when direct substitution results in an indeterminate form (\(\frac{0}{0}\), \(\frac{\infty}{\infty}\), \(0 \times \infty\), \(\infty - \infty\), \(1^\infty\), \(0^0\), \(\infty^0\)).

In this specific problem involving trigonometric functions, using the substitution \(t = \tan(x/2)\) is another valid approach for calculating the limit, as shown in the scratchpad. As \(x \to \pi\), \(x/2 \to \pi/2\), and \(\tan(x/2) \to \infty\). The function transforms into \(\frac{1-t}{1+t}\), and the limit as \(t \to \infty\) is indeed \(-1\).

Understanding continuity allows us to analyze the behavior of functions and is crucial for many applications in science and engineering.