-3
$\begingroup$

I need to find the following limit.

$\mathbf{\lim \limits_{n \to \infty} \frac{3\cdot 2^{2n} + (-1)^{n}}{5\cdot4^{n} + 1}}$

I tried using L'hopital, but can't differentiate $\mathbf{(-1)^{n}}$.

Could I use Euler's number in this question? I'm not sure how to approach the question. Any help would be appreciated.

  • 0
    Is it the numbers $3.2$ or $5.4$ raised at some power, $3$ times $2$ at some power and $5$ times $4$ at some power? – 2017-02-07
  • 0
    3 times 2, 5 times 4 and not 3.2 or 5.4 – 2017-02-07
  • 0
    Please do not use periods for multiplication – 2017-02-07

2 Answers 2

1

Notice that $2^{2n}=4^n$. Then

$$\lim_{n\to\infty}\frac{3\cdot4^n+(-1)^n}{5\cdot4^n+1}=\lim_{n\to\infty}\frac{\dfrac35\cdot(5\cdot4^n+1-1)+(-1)^n}{5\cdot4^n+1}=\lim_{n\to\infty}\frac35+\lim_{n\to\infty}\frac{-\dfrac35+(-1)^n}{5\cdot4^n+1}=\frac35.$$

1

Hint:

$$\frac{3\cdot2^{2n}+(-1)^n}{5\cdot4^n+1}=\frac{3+\left(-\frac14\right)^n}{5+\left(\frac14\right)^n}$$

As $n\to\infty$, each part goes to zero, leaving...