In this lesson, we will now be solving for limits. However, in order for us to simplify the solutions, we must first look at the different basic properties of limits. Below are some of the commonly used properties.
$\thrm{Property 1:}$ The limit of any constant is equal to the constant.
$$\begin{align} \lim_{x \to a} {c} = c \end{align}$$
$\thrm{Property 2:}$ The limit of the product of a constant and a function is equal to the product of the constant and the limit of the function.
$$\begin{align} \lim_{x \to a} {c \,f(x)}= c \lim_{x \to a} f(x) \end{align}$$
$\thrm{Property 3:}$ The limit of the sum or difference of two or more functions is equal to the sum or difference of their limits.
$$\begin{align} \lim_{x \to a} \left[ f(x) + g(x) \right] = \lim_{x \to a} f(x) + \lim_{x \to a}g(x) \end{align}$$
$\thrm{Property 4:}$ The limit of the product of two or more functions is equal to the product of their limits.
$$\begin{align} \lim_{x \to a} \left[ f(x) \cdot g(x) \right] = \lim_{x \to a}{f(x)} \cdot \lim_{x \to a}{g(x)} \end{align}$$
$\thrm{Property 5:}$ The limit of the quotient of two or more functions is equal to the quotient of their limits, provided the limit of the devisor is not zero.
$$\begin{align} \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x) }{ \lim_{x \to a} g(x) } , \hspace{10pt} where: \, \lim_{x \to a} g(x) \neq 0 \end{align}$$
$\thrm{Property 6:}$ The limit of a power of a function is equal to the power of the limit.
$$\begin{align} \lim_{x \to a} \left[ f(x) \right] ^n =\left[ \lim_{x \to a} f(x) \right]^n \end{align}$$
$\thrm{Property 7:}$ The limit of the root of a function is equal to the root of their limits.
$$\begin{align} \lim_{x \to a}{ \sqrt[n]{f(x)} } = \sqrt[n]{ \lim_{x \to a} {f(x)} } \end{align}$$
$\thrm{Property 8:}$ The limit of $\frac{ \sin \theta }{\theta}$ as $\theta$ approaches zero is equal to 1, given that the angle $\theta$ is in radians.
$$\begin{align} \lim_{\theta \to 0}{\frac{ \sin \theta }{\theta}} =1 , \hspace{10pt} where: \theta \; \text{is in radians} \end{align}$$
$\example{1}$ Evaluate: $ \lim_{x \to 2} \br{x^2+2x+3} $
$\solution$
\[\begin{align*} \lim_{x \to 2} \br{x^2+2x+3} &= \lim_{x \to 2} {x^2} + \lim_{x \to 2} {2x} + \lim_{x \to 2} {3} \\ &= \br{ \,\lim_{x \to 2} {x}}^2 + 2 \br{ \,\lim_{x \to 2} {x} } + 3 \\ &= (2)^2+2(2)+3 \\ &= 11 \tagans \end{align*}\]
$\example{2}$ Evaluate: $ \lim_{t \to 3} \br{t^5+10-2t^2-5t^3} $
$\solution$
\[\begin{align*} \lim_{t \to 3} \br{t^5+10-2t^2-5t^3} &= \lim_{t \to 3} {t^5} + \lim_{t \to 3} {10} - \lim_{t \to 3} {2t^2} - \lim_{t \to 3} {5t^3} \\ &= \br{ \,\lim_{t \to 3} {t}}^5 + 10 - 2 \br{ \,\lim_{t \to 3} {t} }^2 - 5 \br{ \,\lim_{t\to 3} {t} }^3 \\ &= (3)^5+10+2(3)^2+5(3)^3 \\ &= 100 \tagans \end{align*}\]
$\example{3}$ Evaluate: $ \lim_{x \to 1} \br{ \frac{3x^2-2}{x^3+2x^2-x+5} } $
$\solution$
\[\begin{align*} \lim_{x \to 1} \br{ \frac{3x^2-2}{x^3+2x^2-x+5} } &= \frac{ \lim_{x \to 1} \br{3x^2-2} }{ \lim_{x \to 1} \br{x^3+2x^2-x+5} } \\ &= \frac{3(1)^2-2}{(1)^3+2(1)^2-(1)+5} \\ &= \frac{1}{7} \tagans \end{align*}\]
$\example{4}$ Evaluate: $ \lim_{y \to 8} { \left[ \frac{5(2-y)^2}{y^2-4} \right] } $
$\solution$
By Theorem 05, \(\begin{align*} \lim_{y \to 8} { \left[ \frac{5(2-y)^2}{y^2-4} \right] } &= \frac{ \lim_{y \to 8} { \left[ 5(2-y)^2 \right] } }{ \lim_{y \to 8} \br{y^2-4} } \\ &= \frac{5(2-8)^2}{(8)^2-4} \\ &= 3 \tagans \end{align*}\)
$\example{5}$ Evaluate: $ \lim_{x \to 1} \br{ \frac{x^2-1}{x^2+3x-4} } $
$\solution$
Factor both the numerator and the denominator,
\[\begin{align*} \lim_{x \to 1} \br{ \frac{x^2-1}{x^2+3x-4} } &= \lim_{x \to 1} { \left[ \frac{ \cancel{(x-1)}(x+1)}{ \cancel{(x-1)}(x+4)} \right] } \\ &= \lim_{x \to 1} \br{ \frac{x+1}{x+4} } \\ &= \frac{1+1}{1+4} \\ &= \frac{2}{5} \tagans \end{align*}\]$\example{6}$ Evaluate: $ \lim_{y \to 2}{ \sqrt{ \frac{y^3-y^2-y-2}{2y^3-5y^2+5y-6} } } $
$\solution$
Factor both the numerator and the denominator,
\[\begin{align*} \lim_{y \to 2}{ \sqrt{ \frac{y^3-y^2-y-2}{2y^3-5y^2+5y-6} } } &= \lim_{y \to 2}{ \sqrt{ \frac{ \cancel{(y-2)}(y^2+y+1)}{ \cancel{(y-2)}(2y^2-y+3)} } } \\ &= \sqrt{ \frac{(2)^2+2+1}{2(2)^2-2+3} } \\ &= \sqrt{ \frac{7}{9} } \\ &= \frac{ \sqrt{7} }{3} \tagans \end{align*}\]$\example{7}$ Evaluate: $ \lim_{\phi \to 0}{\frac{ \sin 5\phi }{5\phi }} $
$\solution$
$\text{Let:} \, \theta = 5\phi $
$\text{If:} \, \phi = 0, \; \text{then,} \, \theta=0$
Thus,
\[\begin{align*} \lim_{\phi \to 0}{\frac{ \sin 5\phi }{5\phi }} &= \lim_{\theta \to 0}{\frac{ \sin \theta}{\theta }} \\ &=1 \tagans \end{align*}\]$\example{8}$ Evaluate: $ \lim_{x \to 0}{\frac{ \tan x}{x}} $
$\solution$
\[\begin{align*} \lim_{x \to 0}{\frac{ \tan x}{x}} &= \lim_{x \to 0}{\frac{ \sin x}{x \cos x}} \\ &= \lim_{x \to 0}{\frac{ \sin x}{x} \cdot \frac{1}{\cos x}} \\ &= \lim_{x \to 0}{\frac{ \sin x}{x} \cdot \lim_{x \to 0}\frac{1}{\cos x}} \\ &= (1)(1) \\ &=1 \tagans \end{align*}\]