Let the function f(x)=cases_e(1+5x)-_e(1+ x)x, & x010, & x=0cases be continuous at x=0.

Question

Let the function $f(x)=\begin{cases}\dfrac{\log_e(1+5x)-\log_e(1+\alpha x)}{x}, & x
e0\10, & x=0\end{cases}$ be continuous at $x=0$. Then $\alpha$ is equal to

Accepted Answer

$$f(0)=\lim_{x\to0}\frac{\ln(1+5x)}{5x}\cdot5-\frac{\ln(1+\alpha x)}{\alpha x}\cdot\alpha=5-\alpha$$
For continuity, $5-\alpha=10\Rightarrow\alpha=-5$.

Solutions · 1