Since \(\displaystyle \tiny { f(-2) = undefined}\) there is a possible vertical asymptote at \(\displaystyle x=-2\).

As we approach \(\displaystyle x=-2\) from the left, the graph should tend to \(\displaystyle \tiny { +\infty}\). Approaching \(\displaystyle x=-2\) from the right, the graph tends to \(\displaystyle \tiny {-\infty}\).

The only graph that does so is  .

We need to look at the behavior of the function as it tends to \(\displaystyle 2\) from the left.  Therefore the answer is \(\displaystyle \tiny { -\infty}\). 

The limit does not exist because the one-sided limits are not equal; \(\displaystyle \lim_{x \rightarrow -2^-}f(x) = -\infty\), whereas  \(\displaystyle \lim_{x \rightarrow -2^+}f(x) = +\infty\). 

The expression \(\displaystyle {}\lim_{x \to 5^-} \left |-5 \right |\) indicates that all points on the domain are equal to 5 since the absolute value negates negative values.

The \(\displaystyle \lim_{x \to 5^-}\) is to search for the limit as the graph approaches \(\displaystyle x=5\) from the left side of the graph.  Since the absolute value of negative five is five, the graph approaches five from the left.

The correct answer is \(\displaystyle 5\).

This limit can be solved using simple manipulation of the expression inside the limit:

\(\displaystyle \\ \small \small \lim_{x\to 9} \frac{x-9}{\sqrt{x}-3}\\ \\=\lim_{x\to 9} \frac{x-9}{\sqrt{x}-3}\cdot \frac{\sqrt{x}+3}{\sqrt{x}+3}\\ \\=\lim_{x\to 9} \frac{x-9}{x-9}\cdot (\sqrt{x}+3)\\ \\=\lim_{x\to 9} \sqrt{x}+3\\ \\=\sqrt{9}+3=6\)

Consider the domain of the function. Because this equation is a polynomial, x is not restricted by any value. Thus the way to evaluate this limit would simply be to plug the value that x is approaching into the limit equation.

\(\displaystyle \lim_{x \to 3} x^2-2x+6=3^2-2*3+6=9-6+6=9\)

Consider the domain of the function. Because this equation is a polynomial, x is not restricted by any value. Thus the way to evaluate this limit would simply be to plug the value that x is approaching into the limit equation.

\(\displaystyle \lim_{x \to -2}\sqrt{x^4+3x+6}=\sqrt{(-2)^4+3(-2)+6}=\sqrt{16-6+6}=\sqrt{16}=4\)

The limiting situation in this equation would be the denominator. Plug the value that x is approaching into the denominator to see if the denominator will equal 0. In this question, the denominator will not equal zero when x=-2; so we proceed to insert the value of x into the entire equation.

\(\displaystyle \lim_{x \to -2}\frac{x^4-2}{2x^2-3x+2}=\frac{(-2)^4-2}{2(-2)^2-3*(-2)+2}=\frac{16-2}{8+6+2}=\frac{14}{16}=\frac{7}{8}\)

The limiting situation in this equation would be the denominator. Plug the value that x is approaching into the denominator to see if the denominator will equal 0. In this question, the denominator will not equal zero when x=0; so we proceed to insert the value of x into the entire equation.

\(\displaystyle \lim_{x \to 0}\frac{\cos^4 (x)}{5+2x^3}=\frac{\cos^4 (0)}{5+2(0)^3}=\frac{1}{5}\)

The limiting situation in this equation would be the denominator. Plug the value that x is approaching into the denominator to see if the denominator will equal 0. In this question, the denominator will equal zero when x=5; so we try to eliminate the denominator by factoring. When the denominator is no longer zero, we may continue to insert the value of x into the remaining equation.

\(\displaystyle \lim_{x \to 5}\frac{x^2-6x+5}{x-5}=\lim_{x \to 5}\frac{(x-5)(x-1)}{x-5}=\lim_{x \to 5}(x-1)=5-1=4\)