I assume you're supposed to find the limit as n approaches infinity, not x.
You have
[tex]\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}}[/tex]
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
[tex]\displaystyle \lim_{n\to\infty}\frac{4^n-5\times3^n+1}{2\times4^n+2}[/tex]
The same algebraic manipulation gives us
[tex]\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}}[/tex]
Again the exponential terms converge to 0, but this time we're left with the limit 1/2.
