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.

- 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. |

- 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). |

- 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. |

- 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. |

- 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. |

- 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. |

- 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. |

- 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. |

- 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). |