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 directly:

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

✔ Answer: $-1$

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

We notice that at $x = 4$, the numerator becomes $\sqrt{4} - 2 = 2 - 2 = 0$, and denominator is $4 - 4 = 0$. So it's an 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 $x - 4$ (since $x \ne 4$):

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

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 $\frac{0}{0}$ at $x = 0$. Use conjugate multiplication:

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

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

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

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

Now take the limit:

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

✔ 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,

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

For $x \ne -2$, cancel $x + 2$:

$$
= x^2 - 2x + 4
$$

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:

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

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 plugging in $x = -1$: numerator = $2(1) +1 -3 = 2 +1 -3 = 0$, denominator = 0 → indeterminate.

Factor numerator: $2x^2 - x - 3$

Use factoring: look for two numbers that multiply to $2 \cdot (-3) = -6$, add to $-1$: $-3$ and $2$

Split middle term:

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

So:

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

Now take 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 both:

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

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

So:

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

Now take 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 is a difference quotient — likely leading to derivative of $f(x) = \frac{1}{x}$ at $x = 2$, but let's simplify algebraically.

Combine the numerator:

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

Now divide by $x$:

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

Now take 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:

$$
\frac{2h}{h} = 2, \quad h \ne 0
$$

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. $2$

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