Question

Consider the function \(f: \mathbb{R} \to \mathbb{R}\) defined as follows:

\[ f(x) = \begin{cases} c_1 e^{x} - c_2 \log_e\left(\dfrac{1}{x}\right), & \text{if } x > 0 \\ 3, & \text{otherwise} \end{cases} \]

where \(c_1, c_2 \in \mathbb{R}\).

If \(f\) is continuous at \(x = 0\), then \(c_1 + c_2 = \) __________. (answer in integer)

Hint:

Since \(\log_e(1/x) \to +\infty\) as \(x \to 0^{+}\), its coefficient must be zero for the limit to stay finite; then match the remaining term to \(f(0) = 3\).

Answer 1

Correct Answer: 3

Step 1: Write down the continuity condition.

For \(f\) to be continuous at \(x = 0\), the right-hand limit of \(f\) as \(x \to 0^{+}\) must exist and equal \(f(0)\). Since \(f(x) = 3\) for \(x \le 0\), we have \(f(0) = 3\), and the left-hand limit as \(x \to 0^{-}\) is automatically 3 as well because the constant branch already covers \(x \le 0\). So the only condition to enforce is \(\lim_{x \to 0^{+}} f(x) = 3\).

Step 2: Examine the behaviour of the logarithmic term as \(x \to 0^{+}\).

As \(x \to 0^{+}\), the quantity \(\dfrac{1}{x} \to +\infty\), so \(\log_e\left(\dfrac{1}{x}\right) \to +\infty\). This term grows without bound.

Step 3: Force the divergent part to vanish.

The full expression as \(x \to 0^{+}\) is \(c_1 e^{x} - c_2 \log_e\left(\dfrac{1}{x}\right)\). The first term \(c_1 e^{x} \to c_1 e^{0} = c_1\), a finite number, but the second term diverges to \(+\infty\) (if \(c_2 > 0\)) or \(-\infty\) (if \(c_2 < 0\)) unless its coefficient is exactly zero. For the overall limit to be finite (which it must be, since it has to equal 3), we require \(c_2 = 0\).

Step 4: Solve for \(c_1\).

With \(c_2 = 0\), the limit reduces to \(\lim_{x \to 0^{+}} c_1 e^{x} = c_1 e^{0} = c_1\). Setting this equal to \(f(0) = 3\) gives \(c_1 = 3\).

Step 5: Compute \(c_1 + c_2\).

\(c_1 + c_2 = 3 + 0 = 3\).

\[ \boxed{c_1 + c_2 = 3} \]