Evaluate the indicated limit algebraically. Change the form of the function where necessary. Please write clearly with description of each step, thank you very much.
3x^2+4.5
________
x^2-1.5

lim
x → ∞

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.

Find the value of f(4) for the function