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COMEDK 2025 Evening Solved Paper

Question : 147 of 180

Marks: +1, -0

\lim\limits_{x \rightarrow 1} \frac{x^4-1}{x-1}=\lim\limits_{x \rightarrow k} \frac{x^3-k^3}{x^2-k^2}

, then the value of K is :

Solution:

Concept:

Both limits are evaluated using algebraic factorization, then equated to solve for

k

Explanation:

First, evaluate the left-hand limit:

\lim\limits_{x\to1}\frac{x^4-1}{x-1} = \lim\limits_{x\to1}\frac{(x-1)(x^3+x^2+x+1)}{x-1}

Cancel

x-1

, then substitute

x=1

to get

1+1+1+1 = 4

Now evaluate the right-hand limit:

\lim\limits_{x\to k}\frac{x^3-k^3}{x^2-k^2} = \lim\limits_{x\to k}\frac{(x-k)(x^2+xk+k^2)}{(x-k)(x+k)}

Cancel

x-k

, then substitute

x=k

to obtain

\frac{k^2+k^2+k^2}{k+k} = \frac{3k^2}{2k} = \frac{3k}{2}

Set the two results equal:

4 = \frac{3k}{2}

Multiply both sides by

2

8 = 3k

Thus

k = \frac{8}{3}

Answer:

k = \frac{8}{3}

(Option D).

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