![Q.10 Let the function \( f: \mathbb{R} \to \mathbb{R} \) be defined by \[ f(x) = \begin{cases} \dfrac{\tan x}{x} & \text{if } x \neq 0; \\ kx & \text{if } x = 0. \end{cases} \] If \( f \) is continuous at \( x = 0 \), then the value of \( k \) must be equal to ____](https://www.letstalkacademy.com/wp-content/uploads/2025/12/slide_50-11.jpg)
Q.10 Let the function f : ℝ → ℝ be defined by
f(x) =
f(x) =
\[
\begin{cases}
\frac{\tan x}{x} & \text{if } x \neq 0, \\
k & \text{if } x = 0.
\end{cases}
\]
If f is continuous at x = 0, then the value of k must be equal to _____.
Question Statement
Let the function \(f: \mathbb{R} \to \mathbb{R}\) be defined by
$$f(x)=\begin{cases}
\dfrac{\tan x}{x}, & x \neq 0 \\
k, & x = 0
\end{cases}$$
If \(f\) is continuous at \(x=0\), find the value of \(k\).
Concept of Continuity
For a function to be continuous at a point \(x = a\), three conditions must hold:
- \(f(a)\) is defined.
- \(\lim_{x \to a} f(x)\) exists.
- \(\lim_{x \to a} f(x) = f(a)\).
In this problem, \(a = 0\) and \(f(0) = k\). Continuity at \(0\) therefore requires
$$\lim_{x \to 0} f(x) = k.$$
Step-by-Step Solution
Step 1: Write the Limit Condition
For \(x \neq 0\), \(f(x) = \dfrac{\tan x}{x}\).
To ensure continuity at \(0\), compute
$$\lim_{x \to 0} \dfrac{\tan x}{x}.$$
If this limit exists and equals \(L\), then \(k\) must be chosen as \(k = L\).
Step 2: Use Standard Trigonometric Limits
Recall the standard limit \(\lim_{x \to 0} \dfrac{\sin x}{x} = 1\).
Express \(\tan x\) as \(\tan x = \dfrac{\sin x}{\cos x}\). Then
$$\dfrac{\tan x}{x} = \dfrac{\sin x}{x} \cdot \dfrac{1}{\cos x}.$$
Now take the limit \(x \to 0\):
$$\lim_{x \to 0} \dfrac{\tan x}{x} = \left( \lim_{x \to 0} \dfrac{\sin x}{x} \right) \cdot \left( \lim_{x \to 0} \dfrac{1}{\cos x} \right).$$
Since \(\lim_{x \to 0} \dfrac{\sin x}{x} = 1\) and \(\cos 0 = 1\), so \(\lim_{x \to 0} \dfrac{1}{\cos x} = \dfrac{1}{1} = 1\).
Therefore
$$\lim_{x \to 0} \dfrac{\tan x}{x} = 1 \cdot 1 = 1.$$
Step 3: Impose the Continuity Condition
Continuity at \(x = 0\) requires
$$\lim_{x \to 0} \dfrac{\tan x}{x} = f(0) = k.$$
Since \(\lim_{x \to 0} \dfrac{\tan x}{x} = 1\), it follows that
Hence, the piecewise function is continuous at \(x = 0\) only when \(k = 1\).
Explanation of Options
Option Analysis
- \(k = 0\): \(f(0) = 0\) while limit = 1. Not continuous.
- \(k = 1\): \(f(0) = 1\) and limit = 1. Continuous.
- \(k = -1\): \(f(0) = -1\) but limit = 1. Not continuous.
- \(k\) does not exist: False, since limit exists (= 1).
Final Answer
\(k = 1\)
