Determine the value for $$ c $$ so that $$ \lim _{x ightarrow 3} f(x) $$ exists. $$ f(x)=\left\{\begin{array}{l}\frac{1}{3} x+c, ext { for } x3 $$

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Determine the value for $$ c $$ so that $$ \lim _{x ightarrow 3} f(x) $$ exists. $$ f(x)=\left\{\begin{array}{l}\frac{1}{3} x+c, 	ext { for } x<3 \ -x+9,\end{array}ight. $$ for $$ x>3 $$

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1. The limit from the left as $$ x \rightarrow 3^- $$ is determined by the expression for $$ x < 3 $$:
   $$
   \lim_{x \rightarrow 3^-} f(x) = \lim_{x \rightarrow 3^-} \left(\frac{1}{3}x + c\right) = \frac{1}{3}\cdot 3 + c = 1 + c.
   $$

2. The limit from the right as $$ x \rightarrow 3^+ $$ is determined by the expression for $$ x > 3 $$:
   $$
   \lim_{x \rightarrow 3^+} f(x) = \lim_{x \rightarrow 3^+} (-x+9) = -3 + 9 = 6.
   $$

3. For the limit $$ \lim_{x \rightarrow 3} f(x) $$ to exist, the left-hand and right-hand limits must be equal:
   $$
   1 + c = 6.
   $$

4. Solving for $$ c $$:
   $$
   c = 6 - 1 = 5.
   $$

Thus, the value of $$ c $$ is $$ 5 $$.
To ensure the limit $$ \lim_{x \rightarrow 3} f(x) $$ exists, set the left-hand limit equal to the right-hand limit. Calculating both: - Left-hand limit: $$ 1 + c = 6 $$ - Right-hand limit: $$ 6 $$ Solving $$ 1 + c = 6 $$ gives $$ c = 5 $$. Therefore, $$ c = 5 $$.

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