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: Use the continuity rule.
If $f$ is continuous at $x=0$, then the limit of $f$ as $x\to0$ must equal $f(0)=\dfrac{7}{4}$.
Step 2: Take the limit of the top and bottom.
We are told $\displaystyle\lim_{x\to0}h(x)=1$. Let $\displaystyle\lim_{x\to0}g(x)=L$. Then \[ \lim_{x\to0}\frac{2h(x)-g(x)}{(h(x)+7)^{2/3}}=\frac{2(1)-L}{(1+7)^{2/3}}. \]
Step 3: Simplify the denominator power.
$8^{2/3}=(2^3)^{2/3}=2^2=4$.
Step 4: Set the limit equal to $f(0)$.
\[ \frac{2-L}{4}=\frac{7}{4}. \]
Step 5: Match the tops.
Since the bottoms are both $4$, we get $2-L=7$.
Step 6: Solve for $L$.
\[ L=2-7=-5. \] \[ \boxed{\lim_{x\to0}g(x)=-5} \]