Question

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 1

Answer:

$\displaystyle \lim_{x\to \infty}\frac{3x^2+4.5}{x^2-1.5}=3$

Step-by-step explanation:

We want to evaluate the limit:

$\displaystyle \lim_{x\to \infty}\frac{3x^2+4.5}{x^2-1.5}$

To do so, we can divide everything by x². So:

$=\displaystyle \lim_{x\to \infty}\frac{3+4.5/x^2}{1-1.5/x^2}$

Now, we can apply direct substitution:

$\Rightarrow \displaystyle \frac{3+4.5/(\infty)^2}{1-1.5/(\infty)^2}$

Any constant value over infinity tends towards 0. Therefore:

$\displaystyle =\frac{3+0}{1+0}=\frac{3}{1}=3$

Hence:

$\displaystyle \lim_{x\to \infty}\frac{3x^2+4.5}{x^2-1.5}=3$

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:

$\displaystyle \Rightarrow \lim_{x\to\infty}\frac{3x^2}{x^2}$

Simplify:

$\displaystyle =\lim_{x\to\infty}3=3$

The limit of a constant is simply the constant.

We acquire the same answer.