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Question:

If lim_{x→1} x⁴/(x-1) = lim_{x→k} (x³ - k³)/(x² - k²), then k=?

4/3

8/3

2/3

none

Solution:

lim_{x→1} x⁴/(x-1) = 4. Applying L'Hopital's rule: lim_{x→1} 4x³/1 = 4
lim_{x→k} (x³ - k³)/(x² - k²) = lim_{x→k} (x-k)(x²+kx+k²)/(x-k)(x+k) = lim_{x→k} (x²+kx+k²)/(x+k) = (k²+k²+k²)/(2k) = 3k²/2k = (3/2)k
Therefore, 4 = (3/2)k
k = 8/3