Solved Unit I Worksheet 2 - Finding Limits Algebraically | Chegg.com - Free Printable

Let's solve each of these limit problems algebraically, using correct notation and reasoning. We’ll go through them one by one.

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1. $\lim_{x \to 2} (-x^2 + x - 2)$

This is a polynomial function, which is continuous everywhere. So we can plug in $x = 2$ directly:

$$
\lim_{x \to 2} (-x^2 + x - 2) = -(2)^2 + 2 - 2 = -4 + 2 - 2 = -4
$$

✔ Answer: $-4$

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2. $\lim_{x \to 1} \cos(\pi x)$

Cosine is continuous everywhere, so we can evaluate at $x = 1$:

$$
\cos(\pi \cdot 1) = \cos(\pi) = -1
$$

✔ Answer: $-1$

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3. $\lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4}$

We notice that as $x \to 4$, both numerator and denominator go to 0 — indeterminate form $\frac{0}{0}$. Let’s rationalize the numerator.

Multiply numerator and denominator by the conjugate: $\sqrt{x} + 2$

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

Cancel $x - 4$ (valid for $x \ne 4$, but since we're taking a limit):

$$
= \frac{1}{\sqrt{x} + 2}
$$

Now take the limit as $x \to 4$:

$$
\lim_{x \to 4} \frac{1}{\sqrt{x} + 2} = \frac{1}{\sqrt{4} + 2} = \frac{1}{2 + 2} = \frac{1}{4}
$$

✔ Answer: $\frac{1}{4}$

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4. $\lim_{x \to 0} \frac{\sqrt{x+3} - \sqrt{3}}{x}$

Again, this is $\frac{0}{0}$ form. Rationalize the numerator:

Multiply numerator and denominator by the conjugate: $\sqrt{x+3} + \sqrt{3}$

$$
\frac{\sqrt{x+3} - \sqrt{3}}{x} \cdot \frac{\sqrt{x+3} + \sqrt{3}}{\sqrt{x+3} + \sqrt{3}} = \frac{(x+3) - 3}{x(\sqrt{x+3} + \sqrt{3})} = \frac{x}{x(\sqrt{x+3} + \sqrt{3})}
$$

Cancel $x$ (for $x \ne 0$):

$$
= \frac{1}{\sqrt{x+3} + \sqrt{3}}
$$

Now take the limit as $x \to 0$:

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

✔ Answer: $\frac{1}{2\sqrt{3}}$

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5. $\lim_{x \to -2} \frac{x^3 + 8}{x + 2}$

Note: $x^3 + 8 = x^3 + 2^3$, which is a sum of cubes:

$$
x^3 + 8 = (x + 2)(x^2 - 2x + 4)
$$

So,

$$
\frac{x^3 + 8}{x + 2} = \frac{(x + 2)(x^2 - 2x + 4)}{x + 2} = x^2 - 2x + 4 \quad \text{(for } x \ne -2\text{)}
$$

Now take the limit:

$$
\lim_{x \to -2} (x^2 - 2x + 4) = (-2)^2 - 2(-2) + 4 = 4 + 4 + 4 = 12
$$

✔ Answer: $12$

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6. $\lim_{x \to 4} \frac{x^2 - 4x}{x^2 - 3x - 4}$

Factor numerator and denominator.

Numerator: $x^2 - 4x = x(x - 4)$

Denominator: $x^2 - 3x - 4 = (x - 4)(x + 1)$

So:

$$
\frac{x(x - 4)}{(x - 4)(x + 1)} = \frac{x}{x + 1} \quad \text{(for } x \ne 4\text{)}
$$

Now take the limit:

$$
\lim_{x \to 4} \frac{x}{x + 1} = \frac{4}{4 + 1} = \frac{4}{5}
$$

✔ Answer: $\frac{4}{5}$

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7. $\lim_{x \to -1} \frac{2x^2 - x - 3}{x + 1}$

Try factoring numerator:

$2x^2 - x - 3$: look for two numbers that multiply to $2 \cdot (-3) = -6$, add to $-1$. Those are $-3$ and $2$.

$$
2x^2 - x - 3 = 2x^2 + 2x - 3x - 3 = 2x(x + 1) -3(x + 1) = (2x - 3)(x + 1)
$$

So:

$$
\frac{(2x - 3)(x + 1)}{x + 1} = 2x - 3 \quad \text{(for } x \ne -1\text{)}
$$

Now take the limit:

$$
\lim_{x \to -1} (2x - 3) = 2(-1) - 3 = -2 - 3 = -5
$$

✔ Answer: $-5$

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8. $\lim_{x \to 2} \frac{x^2 - 4}{x^2 - 5x + 6}$

Factor numerator and denominator.

Numerator: $x^2 - 4 = (x - 2)(x + 2)$

Denominator: $x^2 - 5x + 6 = (x - 2)(x - 3)$

So:

$$
\frac{(x - 2)(x + 2)}{(x - 2)(x - 3)} = \frac{x + 2}{x - 3} \quad \text{(for } x \ne 2\text{)}
$$

Now take the limit:

$$
\lim_{x \to 2} \frac{x + 2}{x - 3} = \frac{2 + 2}{2 - 3} = \frac{4}{-1} = -4
$$

✔ Answer: $-4$

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9. $\lim_{x \to 0} \frac{\frac{1}{2+x} - \frac{1}{2}}{x}$

This looks like a difference quotient. Let’s simplify algebraically.

First, combine the terms in the numerator:

$$
\frac{1}{2+x} - \frac{1}{2} = \frac{2 - (2 + x)}{2(2 + x)} = \frac{-x}{2(2 + x)}
$$

Now divide by $x$:

$$
\frac{-x}{2(2 + x)} \cdot \frac{1}{x} = \frac{-1}{2(2 + x)} \quad \text{(for } x \ne 0\text{)}
$$

Now take the limit as $x \to 0$:

$$
\lim_{x \to 0} \frac{-1}{2(2 + x)} = \frac{-1}{2(2)} = -\frac{1}{4}
$$

✔ Answer: $-\frac{1}{4}$

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10. $\lim_{h \to 0} \frac{2(x + h) - 2x}{h}$

Simplify numerator:

$$
2(x + h) - 2x = 2x + 2h - 2x = 2h
$$

So:

$$
\frac{2h}{h} = 2 \quad \text{(for } h \ne 0\text{)}
$$

Then:

$$
\lim_{h \to 0} 2 = 2
$$

✔ Answer: $2$

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✔ Final Answers:

1. $-4$
2. $-1$
3. $\frac{1}{4}$
4. $\frac{1}{2\sqrt{3}}$
5. $12$
6. $\frac{4}{5}$
7. $-5$
8. $-4$
9. $-\frac{1}{4}$
10. $2$

Let me know if you'd like these explained with graphs or visual intuition!