Question

day 2 - finding limits algebraically.
1.) $limlimits_{x \to 5} 12x$
2.) $limlimits_{x \to 0} \pi$
3.) $limlimits_{x \to 2} 4x$
4.) $limlimits_{x \to 5} (3x^2 - 4x - 1)$
5.) $limlimits_{x \to 0} (5x^3 - 7x^2 + 2^x - 2)$
6.) $limlimits_{y \to -1} (3y^4 - 6y^3 - 2y)$
7.) $limlimits_{x \to 4} \frac{2x - 4}{x - 1}$
8.) $limlimits_{x \to -2} \frac{x^2 + 4x + 4}{x^2}$
9.) $limlimits_{x \to 1} \frac{2x - 2}{x - 1}$
10.) $limlimits_{x \to 4} \frac{x^2 - 16}{x - 4}$
11.) $limlimits_{t \to -2} \frac{t^3 + 8}{t + 2}$
12.) $limlimits_{x \to 2} \frac{x^2 - 4x + 4}{x^2 + x - 6}$
13.) $limlimits_{x \to -1} \frac{x^2 + 6x + 5}{x^2 - 3x - 4}$
14.) $limlimits_{x \to 1} \frac{x^3 + x^2 - 5x + 3}{x^3 - 3x + 2}$
15.) $limlimits_{x \to 3} \frac{x}{x - 3}$
16.) $limlimits_{x \to 5} \frac{x}{x^2 - 25}$
17.) $limlimits_{y \to 6} \frac{y + 6}{y^2 - 36}$
18.) $limlimits_{x \to 1} \frac{2x^2 + x - 3}{x - 1}$

Explanation:

Let's solve these limit problems one by one using the properties of limits (direct substitution, factoring, etc.):

Problem 1: $\boldsymbol{\lim_{x \to 5} 12x}$

Step 1: Use direct substitution (limit of a constant multiple of a function)

The limit of $cx$ as $x \to a$ is $ca$. Here, $c = 12$ and $a = 5$.
$\lim_{x \to 5} 12x = 12 \cdot 5$

Step 2: Calculate the product

$12 \cdot 5 = 60$

Step 1: Use the limit of a constant

The limit of a constant $C$ as $x \to a$ is $C$. Here, $C = \pi$ and $a = 0$.
$\lim_{x \to 0} \pi = \pi$

Step 1: Use direct substitution (limit of a constant multiple of a function)

The limit of $cx$ as $x \to a$ is $ca$. Here, $c = 4$ and $a = 2$.
$\lim_{x \to 2} 4x = 4 \cdot 2$

Step 2: Calculate the product

$4 \cdot 2 = 8$

Answer:

$\boldsymbol{60}$

Problem 2: $\boldsymbol{\lim_{x \to 0} \pi}$