Q. 3 - Calculus | StudyQuestionHub

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Q. 3

Question

Use the preceding two problems and the result of Exercise 22 to calculate the following limit

$\begin{array}{lll} lim_{x \to - 1} x \cdot lim_{x \to 4} x lim_{x \to π} x \\ lim_{x \to 0} x^{2} \cdot lim_{x \to 5} x^{2} x_{x \to - 2} \\ lim_{x \to 0} (2 x - 3) \cdot lim_{x \to 1} (1 - x) \cdot lim_{x \to 3} (3 x + 1) \end{array}$

Step-by-Step Solution

Verified

Answer

The limits have been solved.

1Step 1. Given information

We have to calculate the following limit :

$\begin{array}{l} lim_{x \to - 1} x \cdot lim_{x \to 4} x lim_{x \to π} x \\ lim_{x \to 0} x^{2} \cdot lim_{x \to 5} x^{2} lim_{x \to - 2} x^{2} \\ lim_{x \to 0} (2 x - 3) \cdot lim_{x \to 1} (1 - x) \cdot lim_{x \to 3} (3 x + 1) \end{array}$

2Step 2. Find the limit

$lim_{x \to - 1} x$ can be calculated by substituting -1 for c,

$lim_{x \to - 1} x = - 1$

$lim_{x \to 4} x$ can be calculated by substituting 4 for c,

$lim_{x \to 4} x = 4$

$lim_{x \to π} x$ can be calculated by substituting $lim_{x \to - 1} x$ for c,

$lim_{x \to π} x = π$

$lim_{x \to 0} x^{2}$ can be calculated by substituting 0 for c,

$\begin{matrix} lim_{x \to 0} x^{2} = 0^{2} \\ = 0 \end{matrix}$

$lim_{x \to 5} x^{2}$ can be calculated by substituting 5 for c,

$lim_{x \to π} x = π$

$lim_{x \to - 2} x^{2}$ can be calculated by substituting -2 for c,

$\begin{aligned} lim_{x \to - 2} x^{2} = (- 2)^{2} \\ = 4 \end{aligned}$

$lim_{x \to 0} (2 x - 3)$ can be calculated by substituting 0 for c,

$\begin{aligned} lim_{x \to 0} (2 x - 3) = (2 \cdot 0 - 3)^{2} \\ = 9 \end{aligned}$

$lim_{x \to 1} (1 - x)$ can be calculated by substituting 1 for c,

$\begin{matrix} lim_{x \to 1} (1 - x) = (1 - 1) \\ = 0 \end{matrix}$

$lim_{x \to 3} (3 x + 1)$ can be calculated by substituting 3 for c,

$\begin{matrix} lim_{x \to 3} (3 x + 1) = (3 \cdot 3 + 1) \\ = 9 + 1 \\ = 10 \end{matrix}$

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