The value of limn - -4-5n - 1-2-3n6-3n-1 - 5n is

Lim_n → ∞4·5^n+1 2·3^n6·3^n+1 + 5^n = lim_n → ∞20·5^n 2·3^n18·3^n + 5^n Dividing numerator and denominator by 5^n, we get lim_n → ∞20 2 (35)^n18(35)^n + ...

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Standard XI

Mathematics

Indeterminate Forms

The value of ...

Question

The value of $lim n \to \infty \frac{4 \cdot 5^{n + 1} - 2 \cdot 3^{n}}{6 \cdot 3^{n + 1} + 5^{n}}$ is

- 20

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20

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1

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Solution

The correct option is B $20$
$lim n \to \infty \frac{4 \cdot 5^{n + 1} - 2 \cdot 3^{n}}{6 \cdot 3^{n + 1} + 5^{n}}$

$= lim n \to \infty \frac{20 \cdot 5^{n} - 2 \cdot 3^{n}}{18 \cdot 3^{n} + 5^{n}}$

Dividing numerator and denominator by $5^{n}$ , we get
$lim n \to \infty \frac{20 - 2 {(\frac{3}{5})}^{n}}{18 {(\frac{3}{5})}^{n} + 1} = \frac{20}{1} = 20$

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The value of limn → ∞4⋅5n+1−2⋅3n6⋅3n+1+5n is

The correct option is B 20limn → ∞4⋅5n+1−2⋅3n6⋅3n+1+5n =limn → ∞20⋅5n−2⋅3n18⋅3n+5n Dividing numerator and denominator by 5n, we get limn → ∞20−2(35)n18(35)n+1=201=20

The value of $lim n \to \infty \frac{4 \cdot 5^{n + 1} - 2 \cdot 3^{n}}{6 \cdot 3^{n + 1} + 5^{n}}$ is