## Proofs of the Continuity of Basic Transcendental Functions

Once again, we will need to construct delta-epsilon proofs based on the definition of the limit. Recall that the definition of the two-sided limit is:

 $\lim\limits_{x\to c} f(x)=L$   means that for every   $\epsilon>0$,   there exists a   $\delta>0$,   such that for every $x$, the expression   $0< |x-c|<\delta$   implies   $|f(x)-L|<\epsilon$.

For each proof, we also provide a running commentary.

### Exponential Functions

• If   $b>0$,   then   $\lim\limits_{x\to c}b^x=b^c$.
 If   $b=1$,   a constant function will result, and it is continuous. This proof actually requires three cases. Constant functions were previously shown to be continuous. Now assume   $b>1$.   Suppose   $\epsilon>0$   has been provided. Define   $\epsilon_1=\min\left\{\epsilon,\dfrac{b^c}{2}\right\}$. For the case of a "large" value of $b$, we begin our delta-epsilon proof with an arbitrary epsilon. But to guard against the difficulties arising from having too large an epsilon, we choose a smaller positive epsilon if needed. Define   $\delta=\min\{c-\log_b(b^c-\epsilon_1),\log_b(b^c+\epsilon_1)-c\}$. Because we restricted the size of epsilon, both of these quantities will be defined, and both are positive. Then for all $x$, the expression   $0< |x-c| <\delta$   implies Here is the beginning of our chain of implications. $-\delta< x-c <\delta$, $-c+\log_b(b^c-\epsilon_1) < x-c < \log_b(b^c+\epsilon_1)-c$, $\log_b(b^c-\epsilon_1) < x < \log_b(b^c+\epsilon_1)$, After replacing $\delta$ by the appropriate candidates, we added $c$ to both sides. $b^c-\epsilon_1 < b^x < b^c+\epsilon_1$, Now, we have exponentiated both sides. Since we knew that   $b>1$,   the exponentiation function was an increasing function, and the inequalities retain their previous direction. $-\epsilon_1 < b^x-b^c <\epsilon_1$,   and$|b^x-b^c| < \epsilon_1 \le \epsilon$. We then subtract $b^c$ from each expression, and rewrite the inequality using an absolute value. We also recall the definition of $\epsilon_1$. This proves the statement for   $b>1$. If   $b<1$,   we can define $a=\dfrac{1}{b}$.   Then   $a>1$. For the last case, we convert the "small" base into a problem using a "large" base. Then   $\lim\limits_{x\to c}b^x=\lim\limits_{x\to c}\dfrac{1}{a^x} =\dfrac{1}{\lim\limits_{x\to c}a^x}=\dfrac{1}{a^c}=b^c$ After using the Reciprocal Limit Law, we can evaluate because this limit does have a "large" base. Thus we have established the third case, and the entire theorem.

### Logarithmic Functions

• If   $b>0$   and   $c>0$,   then $\lim\limits_{x\to c}\log_b x=\log_b c$.
 The function   $f(x)=b^x$   has inverse   $f^{-1}(x)=\log_b x$. Logarithmic and exponential functions are inverses. Therefore, the continuity of   $f(x)=b^x$   implies the continuity of its inverse. This is the result of the Inverse Limit Law. The result will hold everywhere in the domain of the logarithm (which is why we have restricted $c$ to positive values only).

### Sine Function

• For all real values of $c$, $\lim\limits_{x\to c}\sin x=\sin c$.
 First, we shall assume that   $-\dfrac{\pi}{2} < c < \dfrac{\pi}{2}$.   This restricts the sine function to that part of its domain from which the inverse function is typically defined. This restriction will allow us to easily use the inverse sine function in our argument. Suppose   $\epsilon>0$   has been provided. Define   $\epsilon_1=\min\left\{\epsilon,\dfrac{1-\sin c}{2},\dfrac{1+\sin c}{2}\right\}$. Our delta-epsilon proof requires an arbitrary positive epsilon. To avoid complications from too large an epsilon, we choose a smaller value if needed. Note that all of the candidates for the subscripted epsilon are positive for values of $c$ in the open interval   $\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$. Define   $\delta=\min\{c-\sin^{-1}(\sin c-\epsilon_1),\sin^{-1}(\sin c+\epsilon_1)-c\}$. Because we restricted epsilon, both of these candidates will be defined and positive. Then for all $x$, the expression   $0< |x-c| <\delta$   implies Here begins our chain of implications. $-\delta< x-c <\delta$, $-c+\sin^{-1}(\sin c-\epsilon_1) < x-c < \sin^{-1}(\sin c+\epsilon_1)-c$, $\sin^{-1}(\sin c-\epsilon_1) < x < \sin^{-1}(\sin c+\epsilon_1)$, We replaced $\delta$ according to its definition, and added $c$ to each expression. $\sin c-\epsilon_1 < \sin x < \sin c+\epsilon_1$, $-\epsilon_1 < \sin x -\sin c < \epsilon_1$,   and$|\sin x-\sin c|<\epsilon_1 \le\epsilon$. Then we take the sine of each expression. Since the sine function is increasing on the interval   $\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$,   the direction of the inequality signs is preserved. Then we subtract   $\sin c$,   rewrite the inequality as a single absolute value, and recall the definition of $\epsilon_1$. This finishes the proof for the interval   $\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$. Now assume   $\dfrac{\pi}{2} < c < \dfrac{3\pi}{2}$. Then   $-\dfrac{\pi}{2} < \pi-c < \dfrac{\pi}{2}$. To establish this statement, multiply both sides of the first inequality by   $-1$,   then add $\pi$. Furthermore,   $\sin(\pi-x)=\sin x$. This equality can be demonstrated by using a Difference Formula for sines. Therefore,   $\lim\limits_{x\to c}\sin x=\lim\limits_{x\to c}\sin(\pi-x)=\sin(\pi-c)=\sin c$. We were able to evaluate the limit in the second expression by substitution, because we had established the location of the argument of the function to be in the interval   $\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$. Since   $\sin(x+2\pi)=\sin x$,   we can extend the continuity to all real values of $c$ that are not odd multiples of $\dfrac{\pi}{2}$. Here, we use the periodicity of the sine function. When   $c=\dfrac{\pi}{2}$,   we can determine (through another delta-epsilon proof) that   $\lim\limits_{x\to c+}\sin x=1$   and   $\lim\limits_{x\to c-}\sin x=1$.   Since   $\sin\dfrac{\pi}{2}=1$,   the continuity at   $c=\dfrac{\pi}{2}$   is established. The one-sided limit from the left would be similar to the earlier delta-epsilon proof. We could then recall   $\sin(\pi-x)=\sin x$   to obtain the limit from the right. When the two one-sided limits have the same value, the two-sided limit also has that value. We then do the same thing for   $c=\dfrac{3\pi}{2}$,   and use periodicity on these two values of $c$ to complete the argument. And having done this, the continuity of the sine function for all real numbers will be established.

### Cosine Function

• For all real values of $c$,     $\lim\limits_{x\to c}\cos x=\cos c$.
 Recall that   $\sin\left(\dfrac{\pi}{2}-x\right)=\cos x$. This identity may be derived using the Difference Formula for the sine function. Therefore,   $\lim\limits_{x\to c}\cos x=\lim\limits_{x\to c}\sin\left(\dfrac{\pi}{2}-x\right)=\sin\left(\dfrac{\pi}{2}-c\right)=\cos c$. Here, we exploit the continuity of the sine function and the Composition Limit Law, to obtain the continuity of the cosine function.

### Tangent Function

• For all values of $c$ except odd multiples of $\dfrac{\pi}{2}$, we have   $\lim\limits_{x\to c}\tan x=\tan c$.
 Recall that   $\tan x=\dfrac{\sin x}{\cos x}$. This Ratio Identity is typically obtained from the definition of the three trigonometric functions involved. Therefore,   $\lim\limits_{x\to c}\tan x= \dfrac{\lim\limits_{x\to c}\sin x}{\lim\limits_{x\to c}\cos x}=\dfrac{\sin c}{\cos c}=\tan c$. Here, we have used the continuity of the sine and cosine functions, together with the Quotient Limit Law.

### Cotangent Function

• For all values of $c$ except even multiples of $\dfrac{\pi}{2}$, we have $\lim\limits_{x\to c}\cot x=\cot c$.
 Recall that   $\cot x=\dfrac{\cos x}{\sin x}$. This Ratio Identity is typically obtained from the definition of the three trigonometric functions involved. Therefore,   $\lim\limits_{x\to c}\cot x= \dfrac{\lim\limits_{x\to c}\cos x}{\lim\limits_{x\to c}\sin x}=\dfrac{\cos c}{\sin c}=\cot c$. Once again, we have used the continuity of the sine and cosine functions, together with the Quotient Limit Law.

### Secant Function

• For all values of $c$ except odd multiples of $\dfrac{\pi}{2}$, we have $\lim\limits_{x\to c}\sec x=\sec c$.
 Recall that   $\sec x=\dfrac{1}{\cos x}$. This Ratio Identity is typically obtained from the definition of the two trigonometric functions involved. Therefore,   $\lim\limits_{x\to c}\sec x=\lim\limits_{x\to c}\dfrac{1}{\cos x}=\dfrac{1}{\cos c}=\sec c$. Here, we have used the continuity of the cosine function, together with the Reciprocal Limit Law.

### Cosecant Function

• For all values of $c$ except even multiples of $\dfrac{\pi}{2}$, we have $\lim\limits_{x\to c}\csc x=\csc c$.
 Recall that   $\csc x=\dfrac{1}{\sin x}$. This Ratio Identity is typically obtained from the definition of the two trigonometric functions involved. Therefore,   $\lim\limits_{x\to c}\csc x=\lim\limits_{x\to c}\dfrac{1}{\sin x}=\dfrac{1}{\sin c}=\csc c$. Here, we have used the continuity of the sine function, together with the Reciprocal Limit Law.

### Inverse Trigonometric Functions

• Whenever   $-1 < c < 1$,   we have   $\lim\limits_{x\to c}\sin^{-1}x=\sin^{-1}c$   and   $\lim\limits_{x\to c}\cos^{-1}x=\cos^{-1}c$.
• For all real values of $c$, we have   $\lim\limits_{x\to c}\tan^{-1}x=\tan^{-1}c$   and   $\lim\limits_{x\to c}\cot^{-1}x=\cot^{-1}c$.
• Whenever   $c<-1$   or   $c>1$,   we have $\lim\limits_{x\to c}\sec^{-1}x=\sec^{-1}c$   and   $\lim\limits_{x\to c}\csc^{-1}x=\csc^{-1}c$.
 Each of these six functions is the inverse of a function whose continuity has already been demonstrated above. In fact, even their name suggests that they are inverses. But care does have to be taken over the interval on which they are inverses of one another. Therefore, over the appropriate intervals, the limits of these six functions will be equal to the values of those functions. Since continuity of the original six functions had been demonstrated, the Inverse Limit Law implies that the inverse functions will also be continuous (except at the endpoints of their domains).