Question

2^n+1 + 3^n+1/2^n+3^n approaches to infinity is equal to
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Answer 1

Answer:

Given:

2n (n→∞) tends to ∞.

+ 3n also (n→∞) tends to ∞

so the sum gets me ∞.

Now (∞)1/∞ : (∞)0=1

I see no other way. Theorem: n1n=1 as n→∞.

Analogous: (2n−3n)1n = (∞−∞)0 = 1 ???

Is it ok to suppose infinity on any integer kn as n goes to infinity

I hope it's help you

Answer 2

Answer:

$\tt{ \displaystyle \lim_{n \to \infty} \dfrac{2^{n+1} + 3^{n+1}}{2^{n} + 3^{n}} = 3}$

Step-by-step explanation:

Given: $\tt{ \displaystyle \lim_{n \to \infty} \dfrac{2^{n+1} + 3^{n+1}}{2^{n} + 3^{n}} }$

$\tt{ = \displaystyle \lim_{n \to \infty} \dfrac{2^{n}.2 + 3^{n}.3}{2^{n} + 3^{n}}}$

$\tt{ = \displaystyle \lim_{n \to \infty} \dfrac{\dfrac{2^{n}}{3^{n}}.2 + 3}{\dfrac{2^{n}}{3^{n}} + 1}}$

$\tt{ = \displaystyle \lim_{n \to \infty} \dfrac{2.(\dfrac{2}{3})^{n} + 3}{(\dfrac{2}{3})^{n} + 1}}$

On putting the limits, we get

$\tt{ = \displaystyle \lim_{n \to \infty} \dfrac{2 \times0 + 3}{0 + 1}}$

$\tt{ = \dfrac{3}{1}}$

$\tt{ = 3 }$