Let fx=8x3+3x and f−1x be the inverse function of f (x), then Ltx→∞x13f−18x−f−1x=

Ltx→∞f−1xx13x=ft As x→α,t→α =Ltt→αtft13=Ltt=αt8t3+3t13=12similarly f−18xx1x,8x=ft =Ltt→αtft813=2.12=1

Let fx=8x3+3x and f−1x be the inverse function of f (x), then Ltx→∞x13f−18x−f−1x=

Ltx→∞f−1xx13x=ft As x→α,t→α =Ltt→αtft13=Ltt=αt8t3+3t13=12similarly f−18xx1x,8x=ft =Ltt→αtft813=2.12=1

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Let $f (x) = 8 x^{3} + 3 x$ and $f^{- 1} (x)$ be the inverse function of f (x), then $\underset{x \to \infty}{L t} \frac{x^{\frac{1}{3}}}{f^{- 1} (8 x) - f^{- 1} (x)} =$

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Detailed Solution

$\underset{x \to \infty}{L t} \frac{f^{- 1} (x)}{x^{\frac{1}{3}}} x = f (t)$ As $x \to α, t \to α$
$= \underset{t \to α}{L t} \frac{t}{{(f (t))}^{\frac{1}{3}}} = \underset{t = α}{L t} \frac{t}{{(8 t^{3} + 3 t)}^{\frac{1}{3}}} = \frac{1}{2}$
similarly $\frac{f^{- 1} (8 x)}{x^{\frac{1}{x}}}, 8 x = f (t)$
$= \underset{t \to α}{L t} \frac{t}{{(\frac{f (t)}{8})}^{\frac{1}{3}}} = 2. \frac{1}{2} = 1$