Question:medium

If \[ \lim_{x \to 5} \frac{x^k - 5^k}{x - 5} = 500 \], then the value of \( k \), where \( k \in \mathbb{N} \) is

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Remember \(\lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}\). This is also the derivative of \(x^n\) at \(x = a\).
Updated On: Jun 4, 2026
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The Correct Option is C

Solution and Explanation

Step 1: Read the limit.
We need the natural number $k$ for which \[ \lim_{x\to 5}\frac{x^k - 5^k}{x - 5} = 500. \]
Step 2: Spot the standard form.
This matches the well known limit \[ \lim_{x\to a}\frac{x^n - a^n}{x - a} = n\,a^{n-1}. \] It is just the definition of the derivative of $x^k$ at the point $x = 5$.
Step 3: Apply the formula.
Here $a = 5$ and $n = k$, so the limit equals \[ k\cdot 5^{k-1}. \]
Step 4: Set it equal to 500.
\[ k\cdot 5^{k-1} = 500. \]
Step 5: Try small natural numbers.
For $k = 3$: $3\cdot 5^{2} = 3\cdot 25 = 75$, too small. For $k = 4$: $4\cdot 5^{3} = 4\cdot 125 = 500$, exactly right.
Step 6: Confirm the answer.
So $k = 4$ gives the value $500$. \[ \boxed{k = 4} \]
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