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

$f(x)=\{\begin{array}{ccc}1-x^3& If& x1\\ ln(x)& If& x>1\end{array}$

On the interval $(-,1),f$ function therefore $r$ in carbinanus an $(-2,1).$

On the tetervat $(1,-\%)$ Til $$ LEgative $2$

At $x=1,$

$\underset{x{1}^{-}}{lim}(x)=\underset{x{1}^{-}}{lim}(,)=$

and

$\underset{x{1}^{}}{lim}(x)=\underset{x{1}^{-}}{lim}()=$

so $\underset{x1}{lim}(x)=$

$$

NotdHotp? $-$ Alwa, $(2)=-$ Also, $(1)=$

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$p$

$q,\underset{1}{lim}(x)=$ e

$q,$