SOLUTION: Basic calculus solved problems - Studypool - Free Printable

Basic Calculus Exam: Part 1 - Limits

Below, I will solve each limit problem step by step. Let's go through them one by one.

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#### 1. $\lim_{x \to 1} \frac{\sqrt[3]{x} - 1}{\sqrt{x} - 1}$

To evaluate this limit, we can use the substitution $ x = t^6 $, so that:
- $\sqrt[3]{x} = t^2$
- $\sqrt{x} = t^3$

As $ x \to 1 $, $ t \to 1 $. The limit becomes:
$$
\lim_{t \to 1} \frac{t^2 - 1}{t^3 - 1}
$$
Factorize the numerator and denominator:
$$
t^2 - 1 = (t - 1)(t + 1), \quad t^3 - 1 = (t - 1)(t^2 + t + 1)
$$
Thus:
$$
\lim_{t \to 1} \frac{(t - 1)(t + 1)}{(t - 1)(t^2 + t + 1)} = \lim_{t \to 1} \frac{t + 1}{t^2 + t + 1}
$$
Substitute $ t = 1 $:
$$
\frac{1 + 1}{1^2 + 1 + 1} = \frac{2}{3}
$$
So:
$$
\boxed{\frac{2}{3}}
$$

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

This is a indeterminate form of type $ \frac{0}{0} $. Rationalize the numerator:
$$
\frac{\sqrt{x + 3} - \sqrt{3}}{x} \cdot \frac{\sqrt{x + 3} + \sqrt{3}}{\sqrt{x + 3} + \sqrt{3}} = \frac{(x + 3) - 3}{x(\sqrt{x + 3} + \sqrt{3})} = \frac{x}{x(\sqrt{x + 3} + \sqrt{3})}
$$
Simplify:
$$
\lim_{x \to 0} \frac{1}{\sqrt{x + 3} + \sqrt{3}}
$$
Substitute $ x = 0 $:
$$
\frac{1}{\sqrt{0 + 3} + \sqrt{3}} = \frac{1}{2\sqrt{3}}
$$
So:
$$
\boxed{\frac{1}{2\sqrt{3}}}
$$

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#### 3. $\lim_{x \to 0} \frac{\cos(2x) - 1}{\cos(x) - 1}$

Use the trigonometric identity $ \cos(2x) = 2\cos^2(x) - 1 $:
$$
\cos(2x) - 1 = 2\cos^2(x) - 2 = 2(\cos^2(x) - 1)
$$
The limit becomes:
$$
\lim_{x \to 0} \frac{2(\cos^2(x) - 1)}{\cos(x) - 1}
$$
Factorize $ \cos^2(x) - 1 = (\cos(x) - 1)(\cos(x) + 1) $:
$$
\lim_{x \to 0} \frac{2(\cos(x) - 1)(\cos(x) + 1)}{\cos(x) - 1} = \lim_{x \to 0} 2(\cos(x) + 1)
$$
Substitute $ x = 0 $:
$$
2(\cos(0) + 1) = 2(1 + 1) = 4
$$
So:
$$
\boxed{4}
$$

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#### 4. $\lim_{x \to 27} \frac{x - 27}{x^{1/3} - 3}$

Let $ y = x^{1/3} $. Then $ x = y^3 $, and as $ x \to 27 $, $ y \to 3 $. The limit becomes:
$$
\lim_{y \to 3} \frac{y^3 - 27}{y - 3}
$$
Factorize $ y^3 - 27 = (y - 3)(y^2 + 3y + 9) $:
$$
\lim_{y \to 3} \frac{(y - 3)(y^2 + 3y + 9)}{y - 3} = \lim_{y \to 3} (y^2 + 3y + 9)
$$
Substitute $ y = 3 $:
$$
3^2 + 3(3) + 9 = 9 + 9 + 9 = 27
$$
So:
$$
\boxed{27}
$$

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#### 5. $\lim_{x \to 0^-} x^3 \cos\left(\frac{2}{x}\right)$

As $ x \to 0^- $, $ x^3 \to 0^- $ and $ \cos\left(\frac{2}{x}\right) $ oscillates between $-1$ and $1$. However, since $ x^3 \to 0 $, the product $ x^3 \cos\left(\frac{2}{x}\right) \to 0 $.
So:
$$
\boxed{0}
$$

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#### 6. $\lim_{x \to -\infty} \frac{x^2 (\sin x + \cos^3 x)}{(x^2 + 1)(x - 3)}$

For large $ |x| $, the dominant terms in the numerator and denominator are $ x^2 \cdot x = x^3 $ and $ x^2 \cdot x = x^3 $, respectively. Thus:
$$
\lim_{x \to -\infty} \frac{x^2 (\sin x + \cos^3 x)}{(x^2 + 1)(x - 3)} \approx \lim_{x \to -\infty} \frac{x^3}{x^3} = \lim_{x \to -\infty} 1 = 1
$$
So:
$$
\boxed{1}
$$

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#### 7. $\lim_{x \to 0} \frac{\sin^{-1}(4x)}{\tan^{-1}(5x)}$

Use the fact that for small $ x $, $ \sin^{-1}(x) \approx x $ and $ \tan^{-1}(x) \approx x $. Thus:
$$
\lim_{x \to 0} \frac{\sin^{-1}(4x)}{\tan^{-1}(5x)} \approx \lim_{x \to 0} \frac{4x}{5x} = \frac{4}{5}
$$
So:
$$
\boxed{\frac{4}{5}}
$$

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#### 8. $\lim_{x \to 0^+} x^x$

Rewrite $ x^x $ as $ e^{x \ln(x)} $. As $ x \to 0^+ $, $ x \ln(x) \to 0 $ (this can be shown using L'Hôpital's Rule on $ \frac{\ln(x)}{1/x} $). Thus:
$$
\lim_{x \to 0^+} x^x = \lim_{x \to 0^+} e^{x \ln(x)} = e^0 = 1
$$
So:
$$
\boxed{1}
$$

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#### 9. $\lim_{x \to +\infty} \frac{7x^2 + x - 100}{2x^2 - 5x}$

Divide the numerator and denominator by $ x^2 $:
$$
\lim_{x \to +\infty} \frac{7 + \frac{1}{x} - \frac{100}{x^2}}{2 - \frac{5}{x}}
$$
As $ x \to +\infty $, $ \frac{1}{x} \to 0 $, $ \frac{100}{x^2} \to 0 $, and $ \frac{5}{x} \to 0 $. Thus:
$$
\lim_{x \to +\infty} \frac{7 + 0 - 0}{2 - 0} = \frac{7}{2}
$$
So:
$$
\boxed{\frac{7}{2}}
$$

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#### 10. $\lim_{x \to +\infty} \sqrt{\frac{x^3 + 7x}{4x^3 + 5}}$

Divide the numerator and denominator inside the square root by $ x^3 $:
$$
\lim_{x \to +\infty} \sqrt{\frac{1 + \frac{7}{x^2}}{4 + \frac{5}{x^3}}}
$$
As $ x \to +\infty $, $ \frac{7}{x^2} \to 0 $ and $ \frac{5}{x^3} \to 0 $. Thus:
$$
\lim_{x \to +\infty} \sqrt{\frac{1 + 0}{4 + 0}} = \sqrt{\frac{1}{4}} = \frac{1}{2}
$$
So:
$$
\boxed{\frac{1}{2}}
$$

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#### 11. $\lim_{x \to +\infty} (3^x + 3^{2x})^{1/x}$

Rewrite the expression:
$$
(3^x + 3^{2x})^{1/x} = \left(3^x(1 + 3^x)\right)^{1/x} = 3 \cdot (1 + 3^x)^{1/x}
$$
As $ x \to +\infty $, $ 3^x \to +\infty $, so $ 1 + 3^x \approx 3^x $. Thus:
$$
(1 + 3^x)^{1/x} \approx (3^x)^{1/x} = 3
$$
So:
$$
\lim_{x \to +\infty} (3^x + 3^{2x})^{1/x} = 3 \cdot 3 = 9
$$
So:
$$
\boxed{9}
$$

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#### 12. $\lim_{x \to -1} f(x)$

From the piecewise definition of $ f(x) $:
- For $ x < -1 $, $ f(x) = \frac{1}{x^2} $.
- As $ x \to -1^- $, $ f(x) \to \frac{1}{(-1)^2} = 1 $.

So:
$$
\boxed{1}
$$

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#### 13. $\lim_{x \to 1} f(x)$

From the piecewise definition of $ f(x) $:
- For $ -1 \leq x < 1 $, $ f(x) = 2 $.
- At $ x = 1 $, $ f(x) = 3 $.
- For $ 1 < x \leq 2 $, $ f(x) = x + 1 $.

The left-hand limit as $ x \to 1^- $ is $ 2 $, and the right-hand limit as $ x \to 1^+ $ is $ 1 + 1 = 2 $. Since both limits are equal, the limit exists and is:
$$
\boxed{2}
$$

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#### 14. $\lim_{x \to 2} f(x)$

From the piecewise definition of $ f(x) $:
- For $ 1 < x \leq 2 $, $ f(x) = x + 1 $.
- As $ x \to 2^- $, $ f(x) \to 2 + 1 = 3 $.
- For $ x > 2 $, $ f(x) = \frac{-1}{(x - 2)^2} $.
- As $ x \to 2^+ $, $ f(x) \to -\infty $.

Since the left-hand limit and right-hand limit are not equal, the limit does not exist. So:
$$
\boxed{\text{DNE}}
$$

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#### 15. $\lim_{x \to \frac{3}{2}} f(x)$

From the piecewise definition of $ f(x) $:
- For $ 1 < x \leq 2 $, $ f(x) = x + 1 $.
- As $ x \to \frac{3}{2} $, $ f(x) \to \frac{3}{2} + 1 = \frac{5}{2} $.

So:
$$
\boxed{\frac{5}{2}}
$$

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Final Answers:

1. $\boxed{\frac{2}{3}}$
2. $\boxed{\frac{1}{2\sqrt{3}}}$
3. $\boxed{4}$
4. $\boxed{27}$
5. $\boxed{0}$
6. $\boxed{1}$
7. $\boxed{\frac{4}{5}}$
8. $\boxed{1}$
9. $\boxed{\frac{7}{2}}$
10. $\boxed{\frac{1}{2}}$
11. $\boxed{9}$
12. $\boxed{1}$
13. $\boxed{2}$
14. $\boxed{\text{DNE}}$
15. $\boxed{\frac{5}{2}}$