Limits Exercise List 02

In this list, we mainly work with indeterminate forms of the type $\dfrac{0}{0}$ and $\dfrac{\infty}{\infty}$. Use factoring, rationalization, and division of the terms by the highest power of $x$.

---

Exercise 1

Calculate:

$$\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$$

View solution

Indeterminate form $\dfrac{0}{0}$. We factor the numerator:

$$\frac{x^2 - 9}{x - 3} = \frac{(x-3)(x+3)}{x-3} = x+3 \quad (x \neq 3)$$

Therefore:

$$\lim_{x \to 3} (x+3) = 6$$

---

Exercise 2

Calculate:

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

View solution

Indeterminate form $\dfrac{0}{0}$. We factor the numerator and denominator:

$$\frac{x^2 - 5x + 6}{x^2 - 4} = \frac{(x-2)(x-3)}{(x-2)(x+2)} = \frac{x-3}{x+2} \quad (x \neq 2)$$

Therefore:

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

---

Exercise 3

Calculate:

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

View solution

Indeterminate form $\dfrac{0}{0}$. We factor:

$$\frac{x^3 - 1}{x^2 - 1} = \frac{(x-1)(x^2 + x + 1)}{(x-1)(x+1)} = \frac{x^2 + x + 1}{x+1} \quad (x \neq 1)$$

Therefore:

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

---

Exercise 4

Calculate:

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

View solution

Indeterminate form $\dfrac{0}{0}$. We rationalize the numerator:

$$\frac{\sqrt{x} - 2}{x - 4} \cdot \frac{\sqrt{x} + 2}{\sqrt{x} + 2} = \frac{x - 4}{(x-4)(\sqrt{x}+2)} = \frac{1}{\sqrt{x}+2}$$

Therefore:

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

---

Exercise 5

Calculate:

$$\lim_{x \to 0} \frac{\sqrt{1+x} - 1}{x}$$

View solution

Indeterminate form $\dfrac{0}{0}$. We rationalize:

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

Therefore:

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

---

Exercise 6

Calculate:

$$\lim_{x \to +\infty} \frac{2x^2 - 3x + 1}{5x^2 + 4x - 7}$$

View solution

Indeterminate form $\dfrac{\infty}{\infty}$. We divide every term by $x^2$ (the highest power):

$$\lim_{x \to +\infty} \frac{2 - \dfrac{3}{x} + \dfrac{1}{x^2}} {5 + \dfrac{4}{x} - \dfrac{7}{x^2}} = \frac{2 - 0 + 0}{5 + 0 - 0} = \frac{2}{5}$$

---

Exercise 7

Calculate:

$$\lim_{x \to +\infty} \frac{3x + 5}{\sqrt{x^2 + 2x}}$$

View solution

Indeterminate form $\dfrac{\infty}{\infty}$. We factor $x$ from the denominator (as $x \to +\infty$, $\sqrt{x^2} = x$):

$$\frac{3x + 5}{\sqrt{x^2 + 2x}} = \frac{3x + 5}{x\sqrt{1 + \dfrac{2}{x}}} = \frac{3 + \dfrac{5}{x}} {\sqrt{1 + \dfrac{2}{x}}}$$

Therefore:

$$\lim_{x \to +\infty} \frac{3 + \dfrac{5}{x}} {\sqrt{1 + \dfrac{2}{x}}} = \frac{3}{1} = 3$$

---

Exercise 8

Calculate:

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

View solution

Now $x \to -\infty$, so $\sqrt{x^2} = |x| = -x$. Thus:

$$\frac{3x + 5}{\sqrt{x^2 + 2x}} = \frac{3x + 5}{-x\sqrt{1 + \dfrac{2}{x}}} = \frac{3 + \dfrac{5}{x}} {-\sqrt{1 + \dfrac{2}{x}}}$$

Therefore:

$$\lim_{x \to -\infty} \frac{3 + \dfrac{5}{x}} {-\sqrt{1 + \dfrac{2}{x}}} = \frac{3}{-1} = -3$$

---

Exercise 9

Calculate:

$$\lim_{x \to 1} \frac{x^2 - \sqrt{x}}{x - 1}$$

View solution

Indeterminate form $\dfrac{0}{0}$. We can factor $x$ from the numerator or use a substitution. One approach is:

$$\frac{x^2 - \sqrt{x}}{x - 1} = \frac{\sqrt{x}(x^{3/2} - 1)}{x - 1}$$

Using $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$ with $a = x^{1/2}$ and $b = 1$:

$$x^{3/2} - 1 = (x^{1/2} - 1)(x + x^{1/2} + 1)$$

And

$$x - 1 = (x^{1/2} - 1)(x^{1/2} + 1).$$

We obtain:

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

Therefore:

$$\lim_{x \to 1} \frac{ \sqrt{x}(x + \sqrt{x} + 1) }{ \sqrt{x} + 1 } = \frac{ 1 \cdot (1 + 1 + 1) }{ 1 + 1 } = \frac{3}{2}$$

---

Exercise 10

Calculate:

$$\lim_{x \to 0} \frac{\sqrt{1+x} - \sqrt{1-x}}{x}$$

View solution

Indeterminate form $\dfrac{0}{0}$. We rationalize the numerator:

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

Therefore:

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