If $\lim_{x \rightarrow 1} \frac{(5x+1)^{1/3}-(x+5)^{1/3}}{(2x+3)^{1/2}-(x+4)^{1/2}}=\frac{m\sqrt{5}}{n(2n)^{2/3}}$, where $\gcd(m,n)=1$, then $8m+12n$ is equal to ____.

JEE Main · 2024 · 04 Apr 2024 · Shift 1 · Limits

lim_{x \to 1} \frac{( 5 x + 1 ) ^{1/3} - ( x + 5 ) ^{1/3}}{( 2 x + 3 ) ^{1/2} - ( x + 4 ) ^{1/2}} = \frac{m 5}{n ( 2 n ) ^{2/3}}

, where

g cd (m, n) = 1

, then

8 m + 12 n

is equal to ____.

Answer: 100

Both numerator and denominator

\to 0

x = 1

; apply L'Hopital.
Numerator derivative at

x = 1

\frac{5}{3} 6^{- 2/3} - \frac{1}{3} 6^{- 2/3} = \frac{4}{3} 6^{- 2/3}

.
Denominator derivative at

x = 1

5^{- 1/2} - \frac{1}{2} 5^{- 1/2} = \frac{1}{2} 5^{- 1/2}

L = \frac{\frac{4}{3} 6 ^{- 2/3}}{\frac{1}{2} 5 ^{- 1/2}} = \frac{8}{3} \cdot \frac{5}{6 ^{2/3}} = \frac{8 5}{3 ( 2 \cdot 3 ) ^{2/3}} .

Comparing with

\frac{m 5}{n ( 2 n ) ^{2/3}}

gives

m = 8, n = 3

(

g cd = 1

8 m + 12 n = 64 + 36 = 100.

Answer: 100.

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