Limits Worksheets :: 25 - 48 - Free Printable

To solve the limits in the worksheet, we will evaluate each limit step by step. Here are the solutions for the given problems:

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(1) \(\lim_{x \to 2} \frac{3x - 6}{x^2 - 4}\)

#### Solution:
- Factor the numerator and denominator:
\[
3x - 6 = 3(x - 2)
\]
\[
x^2 - 4 = (x - 2)(x + 2)
\]
- Substitute these factorizations into the limit:
\[
\lim_{x \to 2} \frac{3(x - 2)}{(x - 2)(x + 2)}
\]
- Cancel the common factor \(x - 2\) (since \(x \neq 2\)):
\[
\lim_{x \to 2} \frac{3}{x + 2}
\]
- Substitute \(x = 2\) into the simplified expression:
\[
\frac{3}{2 + 2} = \frac{3}{4}
\]

Answer:
\[
\boxed{\frac{3}{4}}
\]

---

(2) \(\lim_{x \to 5} \frac{x - 5}{x^2 - 25}\)

#### Solution:
- Factor the denominator:
\[
x^2 - 25 = (x - 5)(x + 5)
\]
- Substitute this factorization into the limit:
\[
\lim_{x \to 5} \frac{x - 5}{(x - 5)(x + 5)}
\]
- Cancel the common factor \(x - 5\) (since \(x \neq 5\)):
\[
\lim_{x \to 5} \frac{1}{x + 5}
\]
- Substitute \(x = 5\) into the simplified expression:
\[
\frac{1}{5 + 5} = \frac{1}{10}
\]

Answer:
\[
\boxed{\frac{1}{10}}
\]

---

(3) \(\lim_{x \to 3} \frac{x^3 - 27}{x - 3}\)

#### Solution:
- Recognize that \(x^3 - 27\) is a difference of cubes:
\[
x^3 - 27 = (x - 3)(x^2 + 3x + 9)
\]
- Substitute this factorization into the limit:
\[
\lim_{x \to 3} \frac{(x - 3)(x^2 + 3x + 9)}{x - 3}
\]
- Cancel the common factor \(x - 3\) (since \(x \neq 3\)):
\[
\lim_{x \to 3} (x^2 + 3x + 9)
\]
- Substitute \(x = 3\) into the simplified expression:
\[
3^2 + 3(3) + 9 = 9 + 9 + 9 = 27
\]

Answer:
\[
\boxed{27}
\]

---

(4) \(\lim_{x \to 0} \frac{\sqrt{x + 1} - 1}{x}\)

#### Solution:
- Rationalize the numerator by multiplying by the conjugate:
\[
\frac{\sqrt{x + 1} - 1}{x} \cdot \frac{\sqrt{x + 1} + 1}{\sqrt{x + 1} + 1} = \frac{(\sqrt{x + 1})^2 - 1^2}{x(\sqrt{x + 1} + 1)} = \frac{x + 1 - 1}{x(\sqrt{x + 1} + 1)} = \frac{x}{x(\sqrt{x + 1} + 1)}
\]
- Cancel the common factor \(x\) (since \(x \neq 0\)):
\[
\lim_{x \to 0} \frac{1}{\sqrt{x + 1} + 1}
\]
- Substitute \(x = 0\) into the simplified expression:
\[
\frac{1}{\sqrt{0 + 1} + 1} = \frac{1}{1 + 1} = \frac{1}{2}
\]

Answer:
\[
\boxed{\frac{1}{2}}
\]

---

(5) \(\lim_{x \to 2} \frac{x^3 + 8}{x^2 - 4}\)

#### Solution:
- Factor the numerator using the sum of cubes:
\[
x^3 + 8 = (x + 2)(x^2 - 2x + 4)
\]
- Factor the denominator using the difference of squares:
\[
x^2 - 4 = (x - 2)(x + 2)
\]
- Substitute these factorizations into the limit:
\[
\lim_{x \to 2} \frac{(x + 2)(x^2 - 2x + 4)}{(x - 2)(x + 2)}
\]
- Cancel the common factor \(x + 2\) (since \(x \neq -2\)):
\[
\lim_{x \to 2} \frac{x^2 - 2x + 4}{x - 2}
\]
- Substitute \(x = 2\) into the simplified expression:
\[
\frac{2^2 - 2(2) + 4}{2 - 2} = \frac{4 - 4 + 4}{0}
\]
This results in an indeterminate form \(\frac{4}{0}\), which indicates the limit does not exist.

Answer:
\[
\boxed{\text{DNE}}
\]

---

(6) \(\lim_{x \to 0} \left[ \frac{1}{x^2} - \frac{1}{x} \right]\)

#### Solution:
- Combine the fractions under a common denominator:
\[
\frac{1}{x^2} - \frac{1}{x} = \frac{1 - x}{x^2}
\]
- Substitute this into the limit:
\[
\lim_{x \to 0} \frac{1 - x}{x^2}
\]
- As \(x \to 0\), the numerator \(1 - x \to 1\) and the denominator \(x^2 \to 0^+\) (since \(x^2\) is always positive):
\[
\frac{1 - x}{x^2} \to \frac{1}{0^+} = +\infty
\]

Answer:
\[
\boxed{+\infty}
\]

---

(7) \(\lim_{x \to 1} \frac{x^2 + 4x - 5}{x^2 - 1}\)

#### Solution:
- Factor the numerator:
\[
x^2 + 4x - 5 = (x + 5)(x - 1)
\]
- Factor the denominator using the difference of squares:
\[
x^2 - 1 = (x - 1)(x + 1)
\]
- Substitute these factorizations into the limit:
\[
\lim_{x \to 1} \frac{(x + 5)(x - 1)}{(x - 1)(x + 1)}
\]
- Cancel the common factor \(x - 1\) (since \(x \neq 1\)):
\[
\lim_{x \to 1} \frac{x + 5}{x + 1}
\]
- Substitute \(x = 1\) into the simplified expression:
\[
\frac{1 + 5}{1 + 1} = \frac{6}{2} = 3
\]

Answer:
\[
\boxed{3}
\]

---

(8) \(\lim_{x \to 0} \frac{\sin(2x)}{x}\)

#### Solution:
- Use the known limit property \(\lim_{u \to 0} \frac{\sin(u)}{u} = 1\). Let \(u = 2x\), so as \(x \to 0\), \(u \to 0\):
\[
\lim_{x \to 0} \frac{\sin(2x)}{x} = \lim_{x \to 0} \frac{\sin(2x)}{2x} \cdot 2 = 2 \cdot \lim_{u \to 0} \frac{\sin(u)}{u} = 2 \cdot 1 = 2
\]

Answer:
\[
\boxed{2}
\]

---

(9) \(\lim_{x \to 0} \frac{1 - \cos(x)}{x^2}\)

#### Solution:
- Use the trigonometric identity \(1 - \cos(x) = 2 \sin^2\left(\frac{x}{2}\right)\):
\[
\lim_{x \to 0} \frac{1 - \cos(x)}{x^2} = \lim_{x \to 0} \frac{2 \sin^2\left(\frac{x}{2}\right)}{x^2}
\]
- Simplify using the substitution \(u = \frac{x}{2}\), so as \(x \to 0\), \(u \to 0\):
\[
\lim_{x \to 0} \frac{2 \sin^2\left(\frac{x}{2}\right)}{x^2} = \lim_{x \to 0} \frac{2 \sin^2(u)}{(2u)^2} = \lim_{x \to 0} \frac{2 \sin^2(u)}{4u^2} = \frac{1}{2} \lim_{u \to 0} \left(\frac{\sin(u)}{u}\right)^2
\]
- Use the known limit \(\lim_{u \to 0} \frac{\sin(u)}{u} = 1\):
\[
\frac{1}{2} \cdot 1^2 = \frac{1}{2}
\]

Answer:
\[
\boxed{\frac{1}{2}}
\]

---

(10) \(\lim_{x \to 0} \frac{\tan(x)}{x}\)

#### Solution:
- Use the known limit property \(\lim_{x \to 0} \frac{\tan(x)}{x} = 1\).

Answer:
\[
\boxed{1}
\]

---

(11) \(\lim_{x \to 0} \frac{\sin(3x)}{x}\)

#### Solution:
- Use the known limit property \(\lim_{u \to 0} \frac{\sin(u)}{u} = 1\). Let \(u = 3x\), so as \(x \to 0\), \(u \to 0\):
\[
\lim_{x \to 0} \frac{\sin(3x)}{x} = \lim_{x \to 0} \frac{\sin(3x)}{3x} \cdot 3 = 3 \cdot \lim_{u \to 0} \frac{\sin(u)}{u} = 3 \cdot 1 = 3
\]

Answer:
\[
\boxed{3}
\]

---

(12) \(\lim_{x \to 0} \frac{\cos(x) - 1}{x}\)

#### Solution:
- Use the known limit property \(\lim_{x \to 0} \frac{\cos(x) - 1}{x^2} = -\frac{1}{2}\). Multiply and divide by \(x\):
\[
\lim_{x \to 0} \frac{\cos(x) - 1}{x} = \lim_{x \to 0} \left( \frac{\cos(x) - 1}{x^2} \cdot x \right) = \left( \lim_{x \to 0} \frac{\cos(x) - 1}{x^2} \right) \cdot \left( \lim_{x \to 0} x \right) = -\frac{1}{2} \cdot 0 = 0
\]

Answer:
\[
\boxed{0}
\]

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Final Answers:

\[
\boxed{
\begin{array}{ll}
(1) & \frac{3}{4} \\
(2) & \frac{1}{10} \\
(3) & 27 \\
(4) & \frac{1}{2} \\
(5) & \text{DNE} \\
(6) & +\infty \\
(7) & 3 \\
(8) & 2 \\
(9) & \frac{1}{2} \\
(10) & 1 \\
(11) & 3 \\
(12) & 0 \\
\end{array}
}
\]