Question

For $a\neq0$ and $b\neq0$, if the real valued function $f(x) = \frac{\sqrt[4]{625+4x}-5}{\sqrt[4]{625+5bx}-5}$ is continuous at $x=0$, then $f(0) =$

• A. $\frac{4b}{5}$
• B. $\frac{5b}{4}$
• C. $\frac{5}{4b}$
• D. $\frac{4}{5b}$

Hint:

Another way to solve such limits is using the formula $\lim_{x\to a} \frac{x^n-a^n}{x-a} = na^{n-1}$. Let $y=625+4x$. As $x\to 0, y\to 625$. The numerator behaves like $(y^{1/4}-625^{1/4})/(y-625) \times (4x)/x$. This can get complicated. L'Hopital's rule is often the most direct method for these types of rational expressions involving roots.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
For a function to be continuous at a point $x=c$, the value of the function at that point, $f(c)$, must be equal to the limit of the function as $x$ approaches $c$. In this case, $f(0) = \lim_{x\to 0} f(x)$. The given function results in an indeterminate form $\frac{0}{0}$ when $x=0$ is substituted directly, so we need to use methods like L'Hopital's rule or series expansion to evaluate the limit. (Note: The problem as stated in the source document is ambiguous. Based on the options, a plausible intended function is $f(x) = \frac{\sqrt[4]{625+4x}-5}{\sqrt[4]{625+5bx}-5}$. We will solve for this function.)
Step 2: Key Formula or Approach
We will evaluate the limit using L'Hopital's Rule, which states that if $\lim_{x\to c} \frac{g(x)}{h(x)}$ is of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then the limit is equal to $\lim_{x\to c} \frac{g'(x)}{h'(x)}$, provided the latter limit exists.
Step 3: Detailed Explanation
The function is $f(x) = \frac{(625+4x)^{1/4}-5}{(625+5bx)^{1/4}-5}$. For the function to be continuous at $x=0$, $f(0)$ must be equal to the limit of $f(x)$ as $x \to 0$. \[ f(0) = \lim_{x\to 0} \frac{(625+4x)^{1/4}-5}{(625+5bx)^{1/4}-5} \] Substituting $x=0$ gives $\frac{(625)^{1/4}-5}{(625)^{1/4}-5} = \frac{5-5}{5-5} = \frac{0}{0}$, which is an indeterminate form. We apply L'Hopital's Rule. Let $g(x) = (625+4x)^{1/4}-5$ and $h(x) = (625+5bx)^{1/4}-5$. Find the derivatives: \[ g'(x) = \frac{d}{dx}((625+4x)^{1/4}-5) = \frac{1}{4}(625+4x)^{-3/4} \cdot (4) = (625+4x)^{-3/4} \] \[ h'(x) = \frac{d}{dx}((625+5bx)^{1/4}-5) = \frac{1}{4}(625+5bx)^{-3/4} \cdot (5b) = \frac{5b}{4}(625+5bx)^{-3/4} \] Now, find the limit of the ratio of the derivatives: \[ \lim_{x\to 0} \frac{g'(x)}{h'(x)} = \lim_{x\to 0} \frac{(625+4x)^{-3/4}}{\frac{5b}{4}(625+5bx)^{-3/4}} \] Substitute $x=0$: \[ \frac{(625)^{-3/4}}{\frac{5b}{4}(625)^{-3/4}} = \frac{1}{\frac{5b}{4}} = \frac{4}{5b} \] Step 4: Final Answer
The limit of the function as $x \to 0$ is $\frac{4}{5b}$. For continuity, $f(0)$ must be equal to this limit. Therefore, $f(0) = \frac{4}{5b}$.