Question

If the function $f(x)=\begin{cases}\frac{2h(x)-g(x)}{(h(x)+7)^{2/3}}, & x\ne0 \\ \frac{7}{4}, & x=0\end{cases}$ is continuous at $x=0$ and $\lim_{x\rightarrow0}h(x)=1$, then $\lim_{x\rightarrow0}g(x)=$

• A. 3
• B. 4
• C. -5
• D. 7

Hint:

Treat limits of components as fixed values when plugging into basic continuity equations to solve algebraic relations directly.

Answer 1

The Correct Option is C

Step 1: Concept
For a function to be continuous at a specific point $x = c$, the limit of the function as $x$ approaches $c$ must exist and equal the exact value of the function at that point: $\lim_{x\rightarrow c}f(x) = f(c)$.

Step 2: Meaning
Since the function $f(x)$ is continuous at $x = 0$, we know that $\lim_{x\rightarrow0}f(x) = f(0) = \frac{7}{4}$.

Step 3: Analysis
Substitute the limit of the components into the limit expression for $x \neq 0$: $\lim_{x\rightarrow0} \frac{2h(x)-g(x)}{(h(x)+7)^{2/3}} = \frac{2(1) - \lim_{x\rightarrow0}g(x)}{(1+7)^{2/3}} = \frac{2 - \lim_{x\rightarrow0}g(x)}{8^{2/3}}$. Since $8^{2/3} = (2^3)^{2/3} = 2^2 = 4$, the equation simplifies to: $\frac{2 - \lim_{x\rightarrow0}g(x)}{4} = \frac{7}{4}$.

Step 4: Conclusion
Equating the numerators directly yields: $2 - \lim_{x\rightarrow0}g(x) = 7 \implies \lim_{x\rightarrow0}g(x) = 2 - 7 = -5$. This corresponds to option (C).

Final Answer: (C)