Step 1: Condition for 0/0 form:
For \( f(0) \) to be defined via continuity, the limit must exist. Since denominator \( \sqrt[5]{625}-5 = 5-5=0 \), the numerator must be 0.
\( \sqrt[5]{a(625)} - 5 = 0 \implies a(625) = 5^5 = 3125 \implies a = \frac{3125}{625} = 5 \).
Step 2: Apply L'Hospital's Rule:
\[ L = \lim_{x \to 0} \frac{\frac{d}{dx}((a(625+x))^{1/5})}{\frac{d}{dx}((625+bx)^{1/5})} = \lim_{x \to 0} \frac{\frac{1}{5}(a(625+x))^{-4/5} \cdot a}{\frac{1}{5}(625+bx)^{-4/5} \cdot b} \]
Substitute \( x=0, a=5 \):
Numerator derivative: \( \frac{1}{5}(3125)^{-4/5} \cdot 5 = (5^5)^{-4/5} = 5^{-4} = \frac{1}{625} \).
Denominator derivative: \( \frac{1}{5}(625)^{-4/5} \cdot b = \frac{b}{5} (5^4)^{-4/5} = \frac{b}{5} 5^{-3.2} \)? No.
\( (625)^{-4/5} = (5^4)^{-4/5} = 5^{-16/5} \). Wait. The problem likely uses 5th root for 625? No \( \sqrt[5]{625} \) is not integer.
Wait, looking closely at the image, the index might be 4 for the denominator?
Image check: Denominator has index 4. \( \sqrt[4]{625+bx}-5 \).
Then \( \sqrt[4]{625} = 5 \). This works perfectly.
Let's re-calculate with Denom index 4.
Denom derivative: \( \frac{1}{4}(625)^{-3/4} \cdot b = \frac{b}{4} (5^4)^{-3/4} = \frac{b}{4} 5^{-3} = \frac{b}{500} \).
Numerator (index 5): \( \frac{1}{625} \).
Limit \( = \frac{1/625}{b/500} = \frac{500}{625b} = \frac{4}{5b} \).
Step 4: Final Answer:
\( f(0) = \frac{4}{5b} \).