3 years ago

Hey someone pls help pls dont send links

Mathematics

2 answers:

galina1969 [7]3 years ago

8 0

Answer:

help with wat man nothing here

Oduvanchick [21]3 years ago

4 0

Answer:

whats the question

Step-by-step explanation:

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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.

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