Question

Tính giá trị của I = nfty}" alt="\lim_{x \to \infty}" align="absmiddle" class="latex-formula"> ($\frac{4^{n} - 5.3^{n} + 1}{2.4^{n} + 2}$)

Answer 1

I assume you're supposed to find the limit as n approaches infinity, not x.

You have

$\displaystyle \lim_{n\to\infty}\frac{4^n-5.3^n+1}{2.4^n+2} = \lim_{n\to\infty}\frac{\left(\dfrac4{5.3}\right)^n-\left(\dfrac{5.3}{5.3}\right)^n+\dfrac1{5.3^n}}{\left(\dfrac{2.4}{5.3}\right)^n+\dfrac2{5.3^n}} \\\\ = \lim_{n\to\infty}\frac{\left(\dfrac4{5.3}\right)^n-1+\dfrac1{5.3^n}}{\left(\dfrac{2.4}{5.3}\right)^n+\dfrac2{5.3^n}}$

For |x| < 1, we have lim |x|ⁿ = 0 as n goes to infinity. Then each exponential term converges to 0, which leaves us with -1/0. This means the limit is negative infinity.

On the other hand, perhaps you meant to write

$\displaystyle \lim_{n\to\infty}\frac{4^n-5\times3^n+1}{2\times4^n+2}$

The same algebraic manipulation gives us

$\displaystyle\lim_{n\to\infty}\frac{\left(\dfrac44\right)^n-5\left(\dfrac34\right)^n+\dfrac1{4^n}}{2\left(\dfrac44\right)^n+\dfrac2{4^n}} = \lim_{n\to\infty}\frac{1-5\left(\dfrac34\right)^n+\dfrac1{4^n}}{2+\dfrac2{4^n}}$

Again the exponential terms converge to 0, but this time we're left with the limit 1/2.