Limx - 4x3 - 64ln x - 3x2 - 16

The correct option is C 16 lim _x→ 4(x^3+64)ln (x 3)x^2 16=lim _x→ 4x^3+64x+4×ln (x 3)x 4 Applying L'Hospital Rule #160; lim _x→ 4x^3+64x+4×1x 3=16 ...

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Byju's Answer

Standard XII

Mathematics

Differentiation under Integral Sign

limx → 4x3 + ...

Question

$lim x \to 4 \frac{(x^{3} + 64) ln (x - 3)}{x^{2} - 16}$

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Solution

The correct option is C 16
$lim x \to 4 \frac{(x^{3} + 64) ln (x - 3)}{x^{2} - 16} = lim x \to 4 \frac{x^{3} + 64}{x + 4} \times \frac{ln (x - 3)}{x - 4}$
Applying L'Hospital Rule
$lim x \to 4 \frac{x^{3} + 64}{x + 4} \times \frac{1}{x - 3} = 16$

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limx→4(x3+64)ln(x−3)x2−16

The correct option is C 16limx→4(x3+64)ln(x−3)x2−16=limx→4x3+64x+4×ln(x−3)x−4Applying L'Hospital Rule limx→4x3+64x+4×1x−3=16

$lim x \to 4 \frac{(x^{3} + 64) ln (x - 3)}{x^{2} - 16}$

The correct option is C 16
$lim x \to 4 \frac{(x^{3} + 64) ln (x - 3)}{x^{2} - 16} = lim x \to 4 \frac{x^{3} + 64}{x + 4} \times \frac{ln (x - 3)}{x - 4}$
Applying L'Hospital Rule
$lim x \to 4 \frac{x^{3} + 64}{x + 4} \times \frac{1}{x - 3} = 16$