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

Use a substitution to make the divergence explicit. Let $t = \dfrac{1}{x}$, so as $x \to 0^{+}$, we have $t \to +\infty$, and $x = \dfrac{1}{t}$.

• Rewrite the branch for $x>0$ in terms of $t$: $f = c_1 e^{1/t} - c_2 \log_e(t)$.
• As $t \to +\infty$, the exponential piece behaves nicely: $e^{1/t} \to e^{0} = 1$, so $c_1 e^{1/t} \to c_1$, a finite constant.
• The logarithmic piece $\log_e(t) \to +\infty$ grows without any bound, so $-c_2\log_e(t)$ can only stay finite (rather than blow up to plus or minus infinity) if its coefficient is zero, that is $c_2 = 0$.
• With the divergent term eliminated, the limit as $t \to \infty$ (equivalently $x \to 0^{+}$) is simply $c_1$.
• Continuity forces this to equal the other branch's value $f(0) = 3$, so $c_1 = 3$.

Therefore $c_1 + c_2 = 3 + 0 = 3$.

$$\boxed{c_1 + c_2 = 3}$$