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Show that $ f $ is continuous on $ (-\infty, \infty ) $.

$ f(x) = \left\{

\begin{array}{ll}

1 - x^2 & \mbox{if $ x \le 1 $}\\

\ln x & \mbox{if $ x > 1 $}

\end{array} \right.$

$f(x)=\left\{\begin{array}{ll}1-x^{2} & \text { if } x \leq 1 \\ \ln x & \text { if } x>1\end{array}\right.$

By Theorem $5,$ since $f(x)$ equals the polynomial $1-x^{2}$ on $(-\infty, 1], f$ is continuous on $(-\infty, 1]$

By Theorem $7,$ since $f(x)$ equals the logarithm function $\ln x$ on $(1, \infty), f$ is continuous on $(1, \infty)$

At $x=1, \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}\left(1-x^{2}\right)=1-1^{2}=0$ and $\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}} \ln x=\ln 1=0 .$ Thus, $\lim _{x \rightarrow 1} f(x)$ exists and

equals $0 .$ Also, $f(1)=1-1^{2}=0 .$ Thus, $f$ is continuous at $x=1 .$ We conclude that $f$ is continuous on $(-\infty, \infty)$

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consider the piecewise function F of X which is equal to one minus x squared If X is less than or equal to one and Ellen of X if X is greater than one now for F of X which is equal to one minus X squared if x is less than or equal to one since one minus expert is a polynomial, then the function must be continued was over the In thermal negative infinity to one for F of X which is equal to l. Innovex If X is greater than one since the domain of natural log of X, yes, 02 infinity, then this function must be continuous on the interval want infinity. And so the only thing that is not established as continuous is for after B container was at the point X equals one at the show that the function is Contains was at x equals one. We recall the definition of the continuity of a function at a given point. And this definition, it says that a function is continuous at a point X equals C. If the three conditions hold, the first one would be for the function to be defined a day or that exists, the next one would be for the limit of this function as X approaches A to B the find or if it exists and then lastly, for the limit of this function as X approaches A to be equal to the value of the function at the And so using this definition, we have for the first part F of one says is equal to 1 -1 Squared, that's equal to zero. And so we say that F of one exists. And then for the next one to determine if the limit exists As X approaches one. We first have to find the one sided limits. So 1st 1 would be for the limit of the function as X approaches one from the left. Now, in this case we will be using one minus X squared. And so this will give us 1 -1 squared or zero. And for the limit As X approaches one from the right of in this case will be using l innovex Since X approaching one from the right Considers X values that are greater than one. And from here we have Ellen of one which is also equal to zero. Now, since the one sided limits are the same, we say that the limit of the function As X approaches one exists. And for the last condition we see that since the limit as X approaches one of f of X, this is equal to zero and this is also equal to ever won. Then you have shown that the LTD the function as X approaches one Equals the value of the function at one. And by definition the function is continuous at tax equals one more. Over the whole function is continuous over the interval negative infinity to infinity