JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Limits

lim_{x \to 1} \frac{x ^{4} - 1}{x - 1} = lim_{x \to k} \frac{x ^{3} - k ^{3}}{x ^{2} - k ^{2}}

, then

k

Answer: (B)

Solution

The left-hand limit is

lim_{x \to 1} \frac{x ^{4} - 1}{x - 1} = lim_{x \to 1} (x^{3} + x^{2} + x + 1) = 4

. The right-hand limit is

lim_{x \to k} \frac{x ^{3} - k ^{3}}{x ^{2} - k ^{2}} = lim_{x \to k} \frac{( x - k ) ( x ^{2} + x k + k ^{2} )}{( x - k ) ( x + k )} = \frac{3 k ^{2}}{2 k} = \frac{3 k}{2}

. Equating the two,

\frac{3 k}{2} = 4

, so

k = \frac{8}{3}

JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Single correct option

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