Solved 3 Limits at Infinity Evaluate the following limits: | Chegg.com - Free Printable

Let's evaluate each of the given limits at infinity step by step. We'll use algebraic manipulation, especially rationalizing and factoring out dominant terms, to simplify expressions that are indeterminate forms (like ∞ − ∞ or ∞/∞).

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Problem 11:

$$
\lim_{x \to -\infty} \left(x^2 - \sqrt{x^4 + 3x^2}\right)
$$

#### Step 1: Recognize the form
As $ x \to -\infty $, both $ x^2 \to \infty $ and $ \sqrt{x^4 + 3x^2} \to \infty $, so this is an $ \infty - \infty $ indeterminate form.

We rationalize by multiplying numerator and denominator by the conjugate:

$$
x^2 - \sqrt{x^4 + 3x^2} = \frac{(x^2 - \sqrt{x^4 + 3x^2})(x^2 + \sqrt{x^4 + 3x^2})}{x^2 + \sqrt{x^4 + 3x^2}} = \frac{x^4 - (x^4 + 3x^2)}{x^2 + \sqrt{x^4 + 3x^2}}
$$

Simplify numerator:
$$
x^4 - x^4 - 3x^2 = -3x^2
$$

So:
$$
= \frac{-3x^2}{x^2 + \sqrt{x^4 + 3x^2}}
$$

Now factor $ x^2 $ from the denominator. But note: $ x \to -\infty $, so $ x < 0 $. When dealing with square roots, we must be careful with signs.

Note:
- $ x^2 = |x|^2 $
- $ \sqrt{x^4 + 3x^2} = \sqrt{x^4(1 + 3/x^2)} = |x^2|\sqrt{1 + 3/x^2} = x^2\sqrt{1 + 3/x^2} $ since $ x^2 > 0 $

But $ x \to -\infty $, so $ x = -|x| $, but $ x^2 $ is still positive.

So:
$$
\sqrt{x^4 + 3x^2} = x^2 \sqrt{1 + \frac{3}{x^2}} \quad \text{(since } x^2 > 0\text{)}
$$

Thus:
$$
\frac{-3x^2}{x^2 + x^2\sqrt{1 + \frac{3}{x^2}}} = \frac{-3x^2}{x^2(1 + \sqrt{1 + \frac{3}{x^2}})} = \frac{-3}{1 + \sqrt{1 + \frac{3}{x^2}}}
$$

Now take limit as $ x \to -\infty $: $ \frac{3}{x^2} \to 0 $

$$
\lim_{x \to -\infty} \frac{-3}{1 + \sqrt{1 + 0}} = \frac{-3}{1 + 1} = \frac{-3}{2}
$$

✔ Answer: $-\frac{3}{2}$

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Problem 12:

$$
\lim_{x \to -\infty} \frac{x - \sqrt{x^2 + 3x}}{4x + 3}
$$

Again, $ x \to -\infty $. Let’s analyze numerator and denominator.

Numerator: $ x - \sqrt{x^2 + 3x} $ → $ -\infty - \sqrt{\infty} $, but need care.

Let’s factor $ x $ in numerator and denominator.

But since $ x \to -\infty $, $ x < 0 $, so $ \sqrt{x^2} = |x| = -x $

Let’s factor $ x $ from the square root.

Write:
$$
\sqrt{x^2 + 3x} = \sqrt{x^2(1 + 3/x)} = |x|\sqrt{1 + 3/x}
$$

Since $ x < 0 $, $ |x| = -x $, so:
$$
\sqrt{x^2 + 3x} = -x \sqrt{1 + \frac{3}{x}}
$$

Then numerator:
$$
x - (-x)\sqrt{1 + \frac{3}{x}} = x + x\sqrt{1 + \frac{3}{x}} = x\left(1 + \sqrt{1 + \frac{3}{x}}\right)
$$

Denominator: $ 4x + 3 $

So:
$$
\frac{x(1 + \sqrt{1 + \frac{3}{x}})}{4x + 3} = \frac{1 + \sqrt{1 + \frac{3}{x}}}{4 + \frac{3}{x}}
$$

Now take $ x \to -\infty $: $ \frac{3}{x} \to 0 $

So:
$$
\frac{1 + \sqrt{1 + 0}}{4 + 0} = \frac{1 + 1}{4} = \frac{2}{4} = \frac{1}{2}
$$

✔ Answer: $ \frac{1}{2} $

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Problem 13:

$$
\lim_{x \to -\infty} \left(x^3 + \sqrt{x^6 + 6x^3}\right)
$$

Again, $ x \to -\infty $. Note:
- $ x^3 \to -\infty $
- $ \sqrt{x^6 + 6x^3} \to \infty $

But let’s analyze carefully.

Factor inside square root:
$$
\sqrt{x^6 + 6x^3} = \sqrt{x^6(1 + 6/x^3)} = |x^3|\sqrt{1 + 6/x^3}
$$

Since $ x \to -\infty $, $ x^3 < 0 $, so $ |x^3| = -x^3 $

Thus:
$$
\sqrt{x^6 + 6x^3} = -x^3 \sqrt{1 + \frac{6}{x^3}}
$$

Now plug into expression:
$$
x^3 + (-x^3)\sqrt{1 + \frac{6}{x^3}} = x^3 \left(1 - \sqrt{1 + \frac{6}{x^3}}\right)
$$

Now $ x^3 \to -\infty $, and $ \sqrt{1 + \frac{6}{x^3}} \approx 1 + \frac{3}{x^3} $ via binomial approximation.

But let’s use a better approach: rationalize.

Let’s write:
$$
x^3 + \sqrt{x^6 + 6x^3} = x^3 + \sqrt{x^6(1 + 6/x^3)} = x^3 + |x^3|\sqrt{1 + 6/x^3}
$$

Since $ x^3 < 0 $, $ |x^3| = -x^3 $, so:
$$
x^3 + (-x^3)\sqrt{1 + 6/x^3} = x^3(1 - \sqrt{1 + 6/x^3})
$$

Now multiply numerator and denominator by conjugate:

Let $ A = x^3(1 - \sqrt{1 + 6/x^3}) $

Multiply numerator and denominator by $ 1 + \sqrt{1 + 6/x^3} $:

$$
A = x^3 \cdot \frac{(1 - \sqrt{1 + 6/x^3})(1 + \sqrt{1 + 6/x^3})}{1 + \sqrt{1 + 6/x^3}} = x^3 \cdot \frac{1 - (1 + 6/x^3)}{1 + \sqrt{1 + 6/x^3}} = x^3 \cdot \frac{-6/x^3}{1 + \sqrt{1 + 6/x^3}}
$$

Simplify:
$$
= \frac{-6}{1 + \sqrt{1 + 6/x^3}}
$$

Now take $ x \to -\infty $: $ 6/x^3 \to 0 $

So:
$$
\lim_{x \to -\infty} \frac{-6}{1 + \sqrt{1 + 0}} = \frac{-6}{1 + 1} = \frac{-6}{2} = -3
$$

✔ Answer: $-3$

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Problem 14:

$$
\lim_{x \to \infty} \frac{2x^2 + \sqrt{9x^4 + 6x^3}}{2 + 3x - 5x^2}
$$

As $ x \to \infty $, highest degree terms dominate.

Numerator: $ 2x^2 + \sqrt{9x^4 + 6x^3} $

First, simplify $ \sqrt{9x^4 + 6x^3} = \sqrt{x^4(9 + 6/x)} = x^2 \sqrt{9 + 6/x} $

So numerator:
$$
2x^2 + x^2 \sqrt{9 + 6/x} = x^2(2 + \sqrt{9 + 6/x})
$$

Denominator: $ -5x^2 + 3x + 2 $

So overall:
$$
\frac{x^2(2 + \sqrt{9 + 6/x})}{-5x^2 + 3x + 2} = \frac{2 + \sqrt{9 + 6/x}}{-5 + 3/x + 2/x^2}
$$

Take limit as $ x \to \infty $: $ 6/x \to 0 $, $ 3/x \to 0 $, $ 2/x^2 \to 0 $

So:
$$
\frac{2 + \sqrt{9}}{-5} = \frac{2 + 3}{-5} = \frac{5}{-5} = -1
$$

✔ Answer: $-1$

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Problem 15:

$$
\lim_{x \to \infty} (4x - \sqrt{16x^2 + 8x})
$$

Indeterminate form $ \infty - \infty $

Rationalize:
$$
4x - \sqrt{16x^2 + 8x} = \frac{(4x - \sqrt{16x^2 + 8x})(4x + \sqrt{16x^2 + 8x})}{4x + \sqrt{16x^2 + 8x}} = \frac{16x^2 - (16x^2 + 8x)}{4x + \sqrt{16x^2 + 8x}}
$$

Simplify numerator:
$$
16x^2 - 16x^2 - 8x = -8x
$$

So:
$$
\frac{-8x}{4x + \sqrt{16x^2 + 8x}}
$$

Now factor $ x $ from denominator:

Note: $ \sqrt{16x^2 + 8x} = \sqrt{x^2(16 + 8/x)} = |x|\sqrt{16 + 8/x} $

Since $ x \to \infty $, $ |x| = x $, so:
$$
\sqrt{16x^2 + 8x} = x\sqrt{16 + 8/x}
$$

So denominator:
$$
4x + x\sqrt{16 + 8/x} = x(4 + \sqrt{16 + 8/x})
$$

Thus:
$$
\frac{-8x}{x(4 + \sqrt{16 + 8/x})} = \frac{-8}{4 + \sqrt{16 + 8/x}}
$$

Now take $ x \to \infty $: $ 8/x \to 0 $, so:
$$
\frac{-8}{4 + \sqrt{16}} = \frac{-8}{4 + 4} = \frac{-8}{8} = -1
$$

✔ Answer: $-1$

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✔ Final Answers Summary:

11. $ \boxed{-\frac{3}{2}} $
12. $ \boxed{\frac{1}{2}} $
13. $ \boxed{-3} $
14. $ \boxed{-1} $
15. $ \boxed{-1} $

All match the provided answers!