Answer:
[tex]\displaystyle \lim_{x\to \infty}\frac{3x^2+4.5}{x^2-1.5}=3[/tex]
Step-by-step explanation:
We want to evaluate the limit:
[tex]\displaystyle \lim_{x\to \infty}\frac{3x^2+4.5}{x^2-1.5}[/tex]
To do so, we can divide everything by x². So:
[tex]=\displaystyle \lim_{x\to \infty}\frac{3+4.5/x^2}{1-1.5/x^2}[/tex]
Now, we can apply direct substitution:
[tex]\Rightarrow \displaystyle \frac{3+4.5/(\infty)^2}{1-1.5/(\infty)^2}[/tex]
Any constant value over infinity tends towards 0. Therefore:
[tex]\displaystyle =\frac{3+0}{1+0}=\frac{3}{1}=3[/tex]
Hence:
[tex]\displaystyle \lim_{x\to \infty}\frac{3x^2+4.5}{x^2-1.5}=3[/tex]
Alternatively, we can simply consider the biggest term of the numerator and the denominator. The term with the strongest influence in the numerator is 3x², and in the denominator it is x². So:
[tex]\displaystyle \Rightarrow \lim_{x\to\infty}\frac{3x^2}{x^2}[/tex]
Simplify:
[tex]\displaystyle =\lim_{x\to\infty}3=3[/tex]
The limit of a constant is simply the constant.
We acquire the same answer.
