kuta software - infinite calculus evaluating limits worksheet ... - Free Printable

We are given two limits to evaluate as $ x \to -\infty $. We’ll solve them algebraically and also verify using the provided graphs.

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Problem 1:

$$
\lim_{x \to -\infty} \frac{x + 2}{x^2 + x + 1}
$$

#### Algebraic Solution:

This is a rational function. As $ x \to -\infty $, we compare the degrees of the numerator and denominator.

- Numerator degree: 1 (from $ x $)
- Denominator degree: 2 (from $ x^2 $)

Since the degree of the denominator is greater than the degree of the numerator, the limit is:

> 0

✔ *Why?* The denominator grows much faster than the numerator in magnitude as $ |x| \to \infty $, so the fraction shrinks toward zero.

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#### Graphical Verification:

Looking at the graph for Problem 1, as $ x \to -\infty $ (far left on the x-axis), the function values approach the x-axis (i.e., $ y = 0 $). The curve flattens out near the horizontal axis.

✔ So, the graph confirms:
$$
\boxed{0}
$$

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Problem 2:

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

#### Algebraic Solution:

Again, a rational function. Let’s compare degrees:

- Numerator degree: 3 (from $ x^3 $)
- Denominator degree: 2 (from $ x^2 $)

Since the numerator has higher degree, the limit will go to either $ +\infty $ or $ -\infty $, depending on the sign of the leading terms.

Let’s factor out the highest power of $ x $ from numerator and denominator:

$$
\frac{3x^3}{3x^2 - 2} = \frac{3x^3}{x^2(3 - \frac{2}{x^2})} = \frac{3x}{3 - \frac{2}{x^2}}
$$

As $ x \to -\infty $:

- $ \frac{2}{x^2} \to 0 $, so denominator → 3
- Numerator → $ 3x \to -\infty $

So overall:
$$
\frac{3x}{3 - 0} = x \to -\infty
$$

✔ Therefore, the limit is:
$$
\boxed{-\infty}
$$

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#### Graphical Verification:

Looking at the graph for Problem 2, as $ x \to -\infty $ (far left), the function goes downward without bound — it trends to $ -\infty $, which matches our calculation.

✔ Graph confirms:
$$
\boxed{-\infty}
$$

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## ✔ Final Answers:

1) $ \boxed{0} $

2) $ \boxed{-\infty} $