Partial Solutions

There is no solution provided for this question.
1. \[\begin{align*} \lim_{x \rightarrow 5} ~ (2x + 3) &= \lim_{x \to 5} ~ 2x + \lim_{x \to 5} ~ 3 & \text{property 3} \\ &= 2 \lim_{x \rightarrow 5} ~ x + \lim_{x \rightarrow 5} ~ 3 & \text{property 4} \\ &= 2(5) + 3 & \text{properties 2 and 1} \\ &= 13 \end{align*}\]
2. \[\begin{align*} \lim_{x \rightarrow 2} ~ (-x^2 + 3x - 2) &= \lim_{x \to 2} ~ (-x^2) + \lim_{x \to 2} ~ (3x) - \lim_{x \to 2}~2 & \text{property 3} \\ &= -\left( \lim_{x \rightarrow 2} ~ x \right)^2 + 3\lim_{x \rightarrow 2} ~ x - \lim_{x \rightarrow 2} ~ 2 & \text{properties 4 and 7}\\ &= -\left( 2 \right)^2 + 3(2) - 2 & \text{properties 2 and 1} \\ &= 0 \end{align*}\]
3. \[\begin{align*} \lim_{t \rightarrow -1} ~ 5(t - 2)(t - 3) &= 5 \left( \lim_{t \rightarrow -1} ~ (t - 2) \right) \left( \lim_{t \rightarrow -1} ~ (t - 3) \right) & \text{properties 4 and 5} \\ &= 5 \left( \lim_{t \rightarrow -1} ~ t - \lim_{t \rightarrow -1} 2 \right) \left( \lim_{t \rightarrow -1} ~ t - \lim_{t \rightarrow -1} ~ 3 \right) & \text{property 3} \\ &= 5 \left( -1 - 2 \right) \left( -1 - 3 \right) & \text{properties 2 and 1} \\ &= 60 \end{align*}\]
4. \[\begin{align*} \lim_{x \rightarrow 3} ~ \dfrac{x + 3}{x - 2} &= \dfrac{ \displaystyle \lim_{x \rightarrow 3} ~ (x + 3)}{ \displaystyle \lim_{x \rightarrow 3} ~ (x - 2)} & \text{property 6} \\ &= \dfrac{ \displaystyle \lim_{x \rightarrow 3} ~ x + \lim_{x \rightarrow 3} 3}{ \displaystyle \lim_{x \rightarrow 3} ~ x - \lim_{x \rightarrow 3} 2} & \text{property 3} \\ &= \dfrac{3 + 3}{3 - 2} & \text{properties 1 and 2} \\ &= \dfrac{6}{1} \\ &= 6 \end{align*}\]
5. \[\begin{align*} \lim_{x \rightarrow 0} ~ 2(2x - 1)^3 &= 2 \left( 2 \lim_{x \rightarrow 0} ~ x - \lim_{x \rightarrow 0} ~ 1 \right)^3 & \text{properties 4, 3, and 7} \\ &= 2(2(0) - 1)^3 & \text{properties 1 and 2} \\ &= -2 \end{align*}\]
6. \[\begin{align*} \lim_{x \rightarrow -3} ~ (5 - x)^{\frac{4}{3}} &= \left( \lim_{x \rightarrow -3} 5 - \lim_{x \rightarrow -3} x \right)^{\frac{4}{3}} & \text{properties 3 and 7} \\ &= (5 - (-3))^{\frac{4}{3}} & \text{properties 1 and 2} \\ &= 16 \end{align*}\]
There is no solution provided for this question.
\[\begin{align*} \lim_{x \rightarrow 1} \dfrac{[f(x)]^3 + [g(x)]^2}{5 - 2g(x)} &= \dfrac{ \left( \displaystyle \lim_{x \rightarrow 1} ~ f(x) \right)^3 + \left( \displaystyle \lim_{x \rightarrow 1} ~ g(x) \right)^2}{\displaystyle \lim_{x \rightarrow 1} ~ 5 - 2 \lim_{x \rightarrow 1} g(x)} \\ &= \dfrac{(-2)^3 + (3)^2}{5 - 2(3)} \\ &= -1 \end{align*}\]
There is no solution provided for this question.
The points \( (a, 0) \) and \( (b, a) \) lie on the line \( y = f(x) \) and so an equation to the line is
\[ f(x) = \dfrac{6}{b - a} (x - a) \]
Similarly, an equation for the line \( y = g(x) \) is
\[ g(x) = \dfrac{2}{b - a}(x - a) \]
If \( x \neq a \), then \( g(x) \neq 0 \) and \( \dfrac{f(x)}{g(x)} = 3 \). Therefore, \( \displaystyle\lim_{x \to a} ~ \dfrac{f(x)}{g(x)} = 3 \).

This limit can also be evaluated using similar triangles.
There is no solution provided for this question.
There are many possible solutions to this question. One possible function is\[ f(x) = \begin{cases} 1 &\text{if } x \gt 0 \\ -1 &\text{if } x \leq 0 \end{cases} \] Observe that \( f(x)^2 = 1 \) for all \( x \in \mathbb{R} \). Then \( \displaystyle \lim_{x \rightarrow 0} ~ [f(x)]^2 = 1 \). However, by definition of \( f(x) \),\[ \displaystyle \lim_{x \rightarrow 0^-} ~ f(x) = -1 \neq 1 = \lim_{x \rightarrow 0^+} ~ f(x) \] so \( \displaystyle \lim_{x \rightarrow 0} ~ f(x) \) does not exist.