If $f$ and $g$ are continuous at $x=0$ and $g(0)\neq 0$, which of the following is necessarily continuous at $x=0$?
A$\dfrac{f(x)}{g(x)}$
B$\dfrac{1}{f(x)}$ always
C$f(x)$ only if $g(x)$ is differentiable
D$f\circ g$ only if $f$ is discontinuous
Answer & Solution
Correct answer: A. $\dfrac{f(x)}{g(x)}$
By the algebra of continuous functions, sums, products, and quotients are continuous wherever defined. Since $g(0)\neq 0$, $\dfrac{f(x)}{g(x)}$ is defined near $x=0$ and is continuous there. Option B is not always true because it requires $f(0)\neq 0$.
Related questions
The angle made by the tangent of the curve $x = a$ ($t + in t$ cost); $y = a(1 + in t)^2A value of $C$ for which the conclusion of mean value theorem holds for the function $f(x)The value of $f(0)$ so that the function $f(x) = \frac{ qrt{1 + x} - qrt[3]{1 + x}}{x}$ b$\int_{-2}^{2}|1 - x^2|dx =$$\lim_{n\to \infty} um_{r = 1}^{n}1 - \frac{1}{n} e^{r}$ isLet $p(x)$ be a function defined on $R$ such that $p'(x) = p'(1 - x)$, for all $x \in [0, If $\frac{dv}{dx} = y + 3 > 0$ and $y(0) = 2$, then $y(\ln 2)$ is equal toThe equation $y^{2} \mathrm{e}^{xy} = 9 \mathrm{e}^{-3} \cdot x^{2}$ defines $y$ as a diff