Question Statement

Examine the continuity of the function $f(x)$ at $x=0$, where:

$f(x)=\{\begin{array}{ll}\frac{\sin x}{x},& x\ne 0\\ \frac{1}{2},& x=0\end{array}$

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Background and Explanation

To determine if a function is continuous at a point, we check whether the limit of the function as $x$ approaches that point equals the function's value at that point. For piecewise functions, special attention is needed at the boundary points where the function definition changes.

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Solution

First, let's clearly write out the piecewise function:

$f(x)=\{\begin{array}{ll}\frac{\sin x}{x},& x\ne 0\\ \frac{1}{2},& x=0\end{array}$

Step 1: Find the limit as $x$ approaches 0

To check continuity at $x=0$, we evaluate ${\lim }_{x\to 0}f(x)$. For $x\ne 0$, the function is defined as $\frac{\sin x}{x}$.

${\lim }_{x\to 0}f(x)={\lim }_{x\to 0}\frac{\sin x}{x}$

This is a standard trigonometric limit. As $x$ approaches 0, the value of $\frac{\sin x}{x}$ approaches 1.

${\lim }_{x\to 0}\frac{\sin x}{x}=1$

Therefore: ${\lim }_{x\to 0}f(x)=1$

Step 2: Find the function value at $x=0$

From the piecewise definition, when $x=0$: $f(0)=\frac{1}{2}$

Step 3: Compare the limit and function value

For continuity at $x=0$, we require: ${\lim }_{x\to 0}f(x)=f(0)$

However, we found: ${\lim }_{x\to 0}f(x)=1\text{and}f(0)=\frac{1}{2}$

Since $1\ne \frac{1}{2}$: ${\lim }_{x\to 0}f(x)\ne f(0)$

Conclusion:

Therefore, $f(x)$ is discontinuous at $x=0$. This is a removable discontinuity because the limit exists but does not equal the function value at that point.

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Key Formulas or Methods Used

• Continuity Condition: A function $f(x)$ is continuous at $x=a$ if ${\lim }_{x\to a}f(x)=f(a)$
• Standard Trigonometric Limit: ${\lim }_{x\to 0}\frac{\sin x}{x}=1$
• Piecewise Function Analysis: Evaluating limits by considering the appropriate branch of the function definition

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Summary of Steps

1. Identify the function definition at the point of interest ($x=0$) and in its neighborhood
2. Calculate the limit as $x$ approaches 0 using the branch for $x\ne 0$: ${\lim }_{x\to 0}\frac{\sin x}{x}=1$
3. Evaluate the function value at $x=0$ directly from the piecewise definition: $f(0)=\frac{1}{2}$
4. Compare the limit and function value: $1\ne \frac{1}{2}$
5. Conclude: Since ${\lim }_{x\to 0}f(x)\ne f(0)$, the function is discontinuous at $x=0$