Rational functions

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We review basic material for rational functions.

The integration technique of partial fractions is a useful technique when evaluating antiderivatives of rational expressions when the degree of the numerator is less than the degree of the denominator.

In order to use this technique, it is important to be able to perform polynomial long division when the degree of the numerator is greater than the degree of the denominator.

The following questions give practice with polynomial long division.

I understand. I do not understand.

Divide: $\frac {2x^2-3x-5}{x+3} = \answer {2x-9} + \frac {22}{x+3}$

Use long division.

Divide: $\frac {4x^2+6x-1}{2x-1} = \answer {2x+4} + \frac {3}{2x-1}$

Use long division.

The technique of partial fractions requires that we are able to factor the denominator of rational expressions. We then have to look for simpler rational functions whose sum or difference could give the original function. For instance, it would be hard to evaluate: $\int \frac {1}{x^2+x}\d x$ However, it is true that: $\begin{align*} \frac{1}{x^2+x} &= \frac{1}{x(x+1)}\\ &=\frac{1}{x}-\frac{1}{x+1} \end{align*}$ and it is easy to integrate the right-hand side of this expression.

We will study how to obtain the right-hand side of this expression, but a conceptual understanding of this procedure that “undoes” the procedure of finding a common denominator is greatly aided by remembering how to add rational functions.

I understand. I do not understand.

Express as a single (reduced) fraction: $\frac {7}{x-2}+\frac {5}{x} = \frac {\answer {-10+12x}}{\answer {-2x+x^2}}$

Express as a single (reduced) fraction: $\frac {x-7}{x^2-4} + \frac {1}{x-3} - \frac {2}{x-5} = \frac {\answer {-14x^2+75x-109}}{\answer {(-5 + x) (-3 + x) (-4 + x^2)}}$

Express as a single (reduced) fraction: $\frac {2}{x^2+7x+11} + \frac {5}{x^2-9} = \frac {\answer {37 + 35 x + 7 x^2}}{\answer {(-9 + x^2) (11 + 7 x + x^2)}}$

L’Hôpital’s rule is a very useful technique in evaluating the limits that arise while solving problems about improper integrals. The following questions review and give practice using L’Hôpital’s rule.

I understand. I do not understand.

List the steps for computing a limit using L’Hôpital’s rule for computing: $\lim _{x\to a} \frac {f(x)}{g(x)}$

Take the derivative of the numerator and denominator. Compute the limits: $\lim _{x\to a}f(x)$ and $\lim _{x\to a}g(x)$ . Use the quotient rule.

If $\lim _{x\to a}f(x) = 0 = \lim _{x\to a}g(x),$ then

The limit is zero. The limit is of the form $\relax \boldsymbol {\tfrac {0}{0}}$ , and hence is indeterminant. The limit is of the form $\relax \boldsymbol {\tfrac {0}{0}}$ , and we may apply L’Hôpital’s rule.

Now we

Take the derivatives of $f$ and $g$ . Use the quotient rule.

Finally, we are done provided that $\lim _{x\to a} \frac {f'(x)}{g'(x)}$

The limit above exists. The limit above is of the form $\relax \boldsymbol {\tfrac {0}{0}}$ .

If the final limit is of the form $\relax \boldsymbol {\tfrac {0}{0}}$ ,

The limit does not exist. L’Hôpital’s rule, has failed and this limit cannot be evaluated. We can use L’Hôpital’s rule again.

Suppose the limit $\lim _{x\to a} f(x) \cdot g(x)$ has the form $\relax \small \boldsymbol {0\cdot \infty }$ . List the steps for for computing this limit. Select all that apply:

The determinant form $\relax \small \boldsymbol {0\cdot \infty }$ tells us that the limit is zero. The determinant form $\relax \small \boldsymbol {0\cdot \infty }$ tells us that the limit is $1$ . The determinant form $\relax \small \boldsymbol {0\cdot \infty }$ tells us that the limit does not exist. Rewrite as $\lim _{x\to a} \frac {f(x)}{g(x)^{-1}}$ and then use L’Hôptial’s rule. Rewrite as $\lim _{x\to a} \frac {g(x)}{f(x)^{-1}}$ and then use L’Hôptial’s rule.

Suppose the limit $\lim _{x\to a} f(x)^{g(x)}$ has the form $\relax \boldsymbol {1^\infty }$ . List the steps for for computing this limit. Select all that apply:

The determinant form $\relax \boldsymbol {1^\infty }$ tells us that the limit is zero. The determinant form $\relax \boldsymbol {1^\infty }$ tells us that the limit is $1$ . The determinant form $\relax \boldsymbol {1^\infty }$ tells us that the limit does not exist. Consider the limit $\lim _{x\to a} e^{\ln (f(x)^{g(x)})}$ .

Now we may use log rules to write $\ln (f(x)^{g(x)}) = g(x)\ln (f(x)).$ In this case $\lim _{x\to a}$ will be of the form:

$\relax \boldsymbol {\tfrac {\#}{0}}$ $\relax \boldsymbol {\tfrac {0}{0}}$ $\relax \small \boldsymbol {0\cdot \infty }$

Use L’Hôpital’s rule twice to compute: $\lim _{x\to 0} \frac {1-\cos (3x)}{4x^2} = \answer {9/8}$

Use L’Hôpital’s rule to compute: $\lim _{x\to 0^+} x^{13x}= \answer {1}$

Use L’Hôpital’s rule to compute: $\lim _{x\to \infty }\left (1+\frac {a}{x}\right )^x = \answer {e^a}$

Use L’Hôpital’s rule to compute: $\lim _{x\to 0}(7x+\cos (x))^{1/x} = \answer {e^7}$