1.3 Q-1

Question Statement

Evaluate the following limits using the theorem of limits:

$$\lim _{x \rightarrow 3}(2x + 4)$$

$$\lim _{x \rightarrow 1}(3x^2 - 2x + 4)$$

$$\lim _{x \rightarrow 3} \sqrt{x^2 + x + 4}$$

$$\lim _{x \rightarrow 2} x \sqrt{x^2 - 4}$$

$$\lim _{x \rightarrow 2} (\sqrt{x^3 + 1} - \sqrt{x^2 + 5})$$

$$\lim _{x \rightarrow -2} \frac{2x^3 + 5x}{3x - 2}$$

Background and Explanation

The limit theorem states that the limit of a sum, product, or other operations can be calculated by evaluating the individual limits and then combining them. Key rules to keep in mind:

• The limit of a constant is just the constant itself.
• The limit of a polynomial can be found by substituting the value of $x$ directly.
• Limits of square roots or rational functions can also be evaluated by substituting values directly, provided the function is continuous at the point of interest.

Solution

i. $\lim _{x \rightarrow 3}(2x + 4)$

We will use the limit theorem to break this expression into simpler parts:

$$\lim _{x \rightarrow 3}(2x + 4) = \lim _{x \rightarrow 3}(2x) + \lim _{x \rightarrow 3}(4)$$

Since the limit of a constant is the constant itself, we get:

$$= 2 \times \lim _{x \rightarrow 3}(x) + 4$$

Substituting $x = 3$:

$$= 2(3) + 4 = 6 + 4 = 10$$

ii. $\lim _{x \rightarrow 1}(3x^2 - 2x + 4)$

Break it into parts:

$$\lim _{x \rightarrow 1}(3x^2 - 2x + 4) = \lim _{x \rightarrow 1}(3x^2) + \lim _{x \rightarrow 1}(-2x) + \lim _{x \rightarrow 1}(4)$$

Evaluate each term:

$$= 3 \times \lim _{x \rightarrow 1}(x^2) - 2 \times \lim _{x \rightarrow 1}(x) + 4$$

Substituting $x = 1$:

$$= 3(1)^2 - 2(1) + 4 = 3 - 2 + 4 = 5$$

iii. $\lim _{x \rightarrow 3} \sqrt{x^2 + x + 4}$

Substitute $x = 3$ directly into the expression inside the square root:

$$\lim _{x \rightarrow 3} \sqrt{x^2 + x + 4} = \sqrt{(3)^2 + 3 + 4} = \sqrt{9 + 3 + 4} = \sqrt{16} = 4$$

iv. $\lim _{x \rightarrow 2} x \sqrt{x^2 - 4}$

We can break the expression into a product of two limits:

$$\lim _{x \rightarrow 2} x \cdot \lim _{x \rightarrow 2} \sqrt{x^2 - 4}$$

Evaluate each part:

$$= 2 \cdot \sqrt{(2)^2 - 4} = 2 \cdot \sqrt{4 - 4} = 2 \cdot 0 = 0$$

v. $\lim _{x \rightarrow 2} (\sqrt{x^3 + 1} - \sqrt{x^2 + 5})$

Break this into two square roots:

$$\lim _{x \rightarrow 2} \left( \sqrt{x^3 + 1} - \sqrt{x^2 + 5} \right)$$

Evaluate each square root separately:

$$= \sqrt{(2)^3 + 1} - \sqrt{(2)^2 + 5} = \sqrt{8 + 1} - \sqrt{4 + 5} = \sqrt{9} - \sqrt{9} = 3 - 3 = 0$$

vi. $\lim _{x \rightarrow -2} \frac{2x^3 + 5x}{3x - 2}$

Substitute $x = -2$ into the numerator and denominator:

$$\lim _{x \rightarrow -2} \frac{2x^3 + 5x}{3x - 2} = \frac{2(-2)^3 + 5(-2)}{3(-2) - 2}$$

Calculate each part:

$$= \frac{2(-8) + (-10)}{-6 - 2} = \frac{-16 - 10}{-8} = \frac{-26}{-8} = \frac{13}{4}$$

Key Formulas or Methods Used

• Limit of a Sum: $\lim _{x \rightarrow a} (f(x) + g(x)) = \lim _{x \rightarrow a} f(x) + \lim _{x \rightarrow a} g(x)$
• Limit of a Product: $\lim _{x \rightarrow a} (f(x) \cdot g(x)) = \lim _{x \rightarrow a} f(x) \cdot \lim _{x \rightarrow a} g(x)$
• Limit of Polynomials and Rational Functions: Direct substitution can be used for continuous functions like polynomials or rational functions.

Summary of Steps

1. For each limit, break down the expression using the limit laws (sum, product, etc.).
2. Evaluate the individual limits by substituting the value of $x$.
3. Simplify the result to get the final answer.