Question

Tính giá trị của I =
`\lim_(x \to \infty)` (
`(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)(4^n-5.3^n+1)/(2.4^n+2) = \lim_(n\to\infty)\frac{\left(\frac4{5.3}\right)^n-\left((5.3)/(5.3)\right)^n+\frac1{5.3^n}}{\left((2.4)/(5.3)\right)^n+\frac2{5.3^n}} \\\\ = \lim_(n\to\infty)\frac{\left(\frac4{5.3}\right)^n-1+\frac1{5.3^n}}{\left((2.4)/(5.3)\right)^n+\frac2{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)(4^n-5*3^n+1)/(2*4^n+2)`

The same algebraic manipulation gives us

`\displaystyle\lim_(n\to\infty)\frac{\left(\frac44\right)^n-5\left(\frac34\right)^n+\frac1{4^n}}{2\left(\frac44\right)^n+\frac2{4^n}} = \lim_(n\to\infty)\frac{1-5\left(\frac34\right)^n+\frac1{4^n}}{2+\frac2{4^n}}`

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