What is the value of 1 ∞ \frac{1}{\infty} ?

Calculus Level 2

lim ⁡ n → ∞ 5 n + 1 + 7 n + 1 5 n − 1 − 7 n − 1 = ? \large\lim_{n\to \infty} \dfrac{5^{n+1}+7^{n+1}}{5^{n-1}-7^{n-1}}=?

0 49 49 − 0.5 -0.5 − 49 -49

Nazmus Sakib by Nazmus sakib , 24, Bangladesh

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1 solution

Chew-Seong Cheong
Nov 15, 2017

L = lim ⁡ x → ∞ 5 n + 1 + 7 n + 1 5 n − 1 − 7 n − 1 Divide up and down by 7 n − 1 . = lim ⁡ x → ∞ 5 2 ( 5 7 ) n − 1 + 7 2 ( 5 7 ) n − 1 − 1 = 0 + 49 0 − 1 = − 49 \begin{aligned} L & = \lim_{x \to \infty} \frac {5^{n+1}+7^{n+1}}{5^{n-1}-7^{n-1}} & \small \color{#3D99F6} \text{Divide up and down by }7^{n-1}. \\ & = \lim_{x \to \infty} \frac {5^2\left(\frac 57\right)^{n-1}+7^2}{\left(\frac 57\right)^{n-1}-1} \\ & = \frac {0+49}{0-1} = \boxed{-49} \end{aligned}

8 Helpful 1 Interesting 1 Brilliant 0 Confused

simple very simple

Ayush Jain - 3 years, 6 months ago

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