First determine one-sided limits. The limit exists if the two one-sided limits have the same value.

The right-hand and the left-hand side limits are equal. Thus, the limit exists and $$ \lim_{x \to 4} f(x) \, \, = \, \, 0 \, . $$

Apply the Third Binomial Formula after multiplying both numerator and denominator by \(\sqrt{1+x} + \sqrt{1-x}\) (recall, this procedure is called rationalizing).

\[ \lim_{x \to 0} \, \frac{\sqrt{1+x} - \sqrt{1-x}}{x} \, \, = \, \, 1 \, . \]

\begin{eqnarray*} \frac{\sqrt{1+x} - \sqrt{1-x}}{x} & = & \frac{\sqrt{1+x} - \sqrt{1-x}}{x} \cdot \frac{\sqrt{1+x} + \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}} \\[1mm] & = & \frac{(1+x) - (1-x)}{x \cdot (\sqrt{1+x} + \sqrt{1-x})} \, \, = \, \, \frac{2x}{x \cdot (\sqrt{1+x} + \sqrt{1-x})}\\[1mm] & = & \frac{2}{\sqrt{1+x} + \sqrt{1-x}} \, \, \stackrel{x \to 0}{\longrightarrow} \, \, \frac{2}{1+1} \, \, = \, \, 1 \, . \end{eqnarray*}

Try to factorize the numerator and check for denominator to be 0.

The graph of \(f\) has vertical asymptote at \(x = 0\).

First, we check if we can factorize the numerator \(x^2 - 5x + 6\). We have
\[ x_{1/2} \, \, = \, \, \frac{5 \pm \sqrt{25-24}}{2} \, \, = \, \, \frac{5 \pm 1}{2} \, \, = \, \left\{ \begin{array}{c} 3 \\ 2 \end{array} \right. \] Thus, \[ \frac{x^2 - 5x + 6}{x^2 - 2x} \, \, = \, \, \frac{(x-3) \cdot (x-2)}{x \cdot (x-2)} \, \, = \, \, \frac{x-3}{x} \qquad \text{for $x \neq 2$} \, , \] such that the graph of \(f\) has a hole at \(x=2\) and a vertical asymptote at \(x = 0\) with (\(x-3 < 0\) for \(x\) close to \(0\))
\[ \lim_{x \to 0^{-}} f(x) \, \, = \, \, \infty \qquad \text{and} \qquad \lim_{x \to 0^{+}} f(x) \, \, = \, \, -\infty \]

• For the function to be continuous, there mustn't be discontinuities or vertical asymptotes.
• To find \(y\) intercept, plug in 0 to the function.
• To find the roots, calculate \(f(x) = 0\).
• Think about what happens to the function when \(x\) is either large positive or large negative.

• Continuity: The function is continuous on the whole \(\mathbb{R}\)
• \(y\)-intercept is -16
• Roots: \(x \in \{ 2, -1, 1 \}\) .
• Limits at infinity: Limits at infinity are both infinity.

• Continuity: The function is a product of powers of polynomials and thus continuous on the whole of \(\mathbb{R}\). I.e. there are no discontinuities and vertical asymptotes.
• \(y\)-intercept: \(f(0) = (-2)^4 (1)^3 (-1) = -16\)
• Roots: Due to the product structure, the roots \(f(x) = 0\) are easily found at \(x \in \{ 2, -1, 1 \}\) \((x-2)^4\) is never negative, hence \(f\) doesn't change sign at \(x=2\) and the graph doesn't cross the \(x\)-axis at \(x = 2\). The graph crosses the \(x\)-axis at \(x = 1\) and \(x = -1\).

• Limits at infinity: When \(x\) is large positive, all three factors are large, so \[ \lim_{x \to \infty} \, \underbrace{\left( x-2 \right)^4}_{\to \, \infty} \underbrace{\left( x+1 \right)^3}_{\to \, \infty} \underbrace{\left( x-1 \right)}_{\to \, \infty} \, \, = \, \, \infty \, . \] When \(x\) is large negative, the first factor is large positive and the second and third factors are both large negative, so \[ \lim_{x \to \infty} \, \underbrace{\left( x-2 \right)^4}_{\to \, \infty} \underbrace{\left( x+1 \right)^3}_{\to \, -\infty} \underbrace{\left( x-1 \right)}_{\to \, -\infty} \, \, = \, \, \infty \, . \]

Thus, we get the following sketch of \( f(x) = \left( x-2 \right)^4 \left( x+1 \right)^3 \left( x-1 \right) \):

Draw the information that you get from the points a.-f. into a common coordinate system. Then, connect the different peices of information. You may want to use auxiliary lines, like for vertical and horizontal asymptotes, that you can better fit the graph to the given information.

We have:

The procedure of settig-up the sketch is similar to that of exercise 4 above.

The function is continuous on \([-1,1]\) if it is continuous for all interior and boundary points.

Step 1: Interior points. If \(-1 < a < 1\) (i.e. in the interior of the interval where both one-sided limits may exist), then using the limit laws, we have \begin{eqnarray*} \lim_{x \to a} f(x) & = & \lim_{x \to a} \left( 1 - \sqrt{1 - x^2} \right) \, \, = \, \, 1 - \lim_{x \to a} \sqrt{1 - x^2}
& = & 1 - \sqrt{ \lim_{x \to a} \left( 1-x^2 \right) } \, \, = \, \, 1 - \sqrt{ 1 - \lim_{x \to a} x^2 }
& = & 1 - \sqrt{ 1 - a^2} \, \, = \, \, f(a) \, . \end{eqnarray*}
Thus, \(f\) is continuous for all \(a \in (-1, 1)\).

Step 2: Boundary points. Similar calculations show that $$ \lim_{x \to -1^+} f(x) \, \, = \, \, 1 \, \, = \, \, f(-1) $$ and $$ \lim_{x \to 1^-} f(x) \, \, = \, \, 1 \, \, = \, \, f(1) $$ so \(f\) is continuous from the right at \(-1\) and continupous from the left at \(1\).

Therefore, \(f\) is continuous on \([-1,1]\).

If either \(f(1)\) or \(f(2)\) is more than 0, while another is less than 0, \(x\) for \(f(x) = 0\) will be between \(1\) and \(2\)

We are looking for a solution of the given equation, that is, a number \(c \in (1, 2)\) such that \(f(c) = 0\). Therefore, we take \(a = 1\), \(b = 2\), and \(N = 0\) in the IVP. We have \[ f(1) \, \, = \, \, 4 - 6 + 3 - 2 \, \, = \, \, -1 \, \, < \, \, 0 \] and \[ f(-1) \, \, = \, \, 32 - 24 + 6 - 2 \, \, = \, \, 12 \, \, > \, \, 0 \, . \] Thus \(f(1) < 0 < f(2)\); i.e. \(N = 0\) is a number between \(f(1)\) and \(f(2)\).
Now \(f\) is continuous since it is a polnomial, so the IVP says there is a number \(c\) between \(1\) and \(2\) such that \(f(c) = 0\).
In other words, the equation \(4 x^3 - 6 x^2 + 3 x - 2 = 0\) has at least one root \(c\) in the interval \( (1, 2) \).