# 1. Determine where f is continuous. f(x) = ln(tan^(-1)x) / (x^2 - 9) 2. Find the limits. (a) lim...

## Question:

1. Determine where f is continuous. {eq}f(x) = \ln \left( \frac{\arctan x}{x^2 - 9} \right) {/eq}

2. Find the limits.

{eq}(a) \lim_{x \rightarrow \infty} \cos(1/x) \\ (b) \lim_{x \rightarrow 0} e^{\sin x} \\ (c) \lim_{x \rightarrow 0} \frac {\tan (7x)}{\sin(3x)} {/eq}

## Limits and Continuity

The idea of limits is central to all of calculus. For example, the derivative is defined using a limit, and the definite integral can be calculated using limits as well. There are several theorems regarding the calculation of limits. One very useful theorem states that if you have the limit of a composition of two continuous functions {eq}f(x) {/eq} and {eq}g(x) {/eq} then the limit can be calculated using {eq}\displaystyle\lim_{x \rightarrow a} f(g(x)) = f(\lim_x \rightarrow a) g(x)) {/eq}. We will use this formula several times in this problem.

Continuity is an equally important idea, due to the fact that a continuous function can be integrated over a closed interval. In order to find intervals over which a function is continuous we first look for any domain restrictions for the function. Then we can study the behavior of the function away from these domain restrictions. A similar theorem holds true for continuous functions as for limits. In particular, the composition of two continuous functions is also continuous. We can use this idea to study the composition of more complicated functions.

1.Determine where {eq}f {/eq} is continuous. {eq}f(x) = \ln \left( \frac{\arctan x}{x^2 - 9} \right) {/eq}

The domain of the natural log function {eq}y = \ln x {/eq} consists of all positive real numbers. Therefore the domain of {eq}f(x) = \ln \left( \frac{\arctan x}{x^2 - 9} \right) {/eq} must satisfy the restriction {eq}\frac{\arctan x}{x^2 - 9} > 0 {/eq}. If {eq}x > 0 {/eq} then {eq}\arctan x > 0 {/eq}. Therefore the denominator must be greater than zero as well. The solution of the inequality {eq}x^2 - 9 > 0 {/eq} with the constraint that {eq}x > 0 {/eq} is {eq}x > 3 {/eq}. There are no other points of discontinuity of this function.

2. Find the limits.

a. {eq}\lim_{x \rightarrow \infty} \cos(1/x) {/eq}

The function {eq}y = \cos (1/x) {/eq} can be considered to be the composition of two functions: {eq}\cos (1/x) = f(g(x)) {/eq} where {eq}f(x) = \cos x {/eq} and {eq}g(x) = \frac{1}{x} {/eq}. Both of these functions are continuous for {eq}x > 0 {/eq}. Therefore we can calculate the limit as follows:

{eq}\begin{eqnarray*} \lim_{x \rightarrow \infty} \cos (1/x) & = & \cos \left( \lim_{x \rightarrow \infty} \frac{1}{x} \right) \\& = & \cos 0 \\&= & 1 \end{eqnarray*} {/eq}

since {eq}\lim_{x \rightarrow \infty} \frac1x = 0. {/eq}

b. {eq}\lim_{x \rightarrow 0} e^{\sin x} {/eq}

The function {eq}y = e^{\sin x} {/eq} can be considered as the composition of two functions: {eq}e^{\sin x} = f(g(x)) {/eq} where {eq}f(x) = e^x {/eq} and {eq}g(x) = \sin x {/eq}. Both of these functions are continuous for all real numbers. Therefore we can calculate the limit as follows:

{eq}\begin{eqnarray*} \lim_{x \rightarrow 0} e^{\sin x} & = & e^{\lim_{x \rightarrow 0} \sin x} \\ & = & e^0 & = & 1 \end{eqnarray*} {/eq}

because of the fact that {eq}\lim_{x \rightarrow 0} \sin x = 0. {/eq}

c. {eq}\lim_{x \rightarrow 0} \frac{\tan (7x)}{\sin(3x)} {/eq}

If we attempt a direct substitution of {eq}x = 0 {/eq} into the function {eq}y = \frac{\tan(7x)}{\sin(3x)} {/eq} we obtain the indeterminant form {eq}0/0 {/eq}. Therefore L'Hopital's Rule can be applied:

{eq}\begin{eqnarray*} \lim_{x \rightarrow 0} \frac{\tan(7x)}{\sin (3x)} & = & \lim_{x \rightarrow 0} \frac{7 \sec^2 (7x)}{3 \cos (3x)} \\ & = & \frac{7 \sec^2 0}{3 \cos 0} \\ & = & \frac73 \end{eqnarray*} {/eq}