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 $$ x>3 $$

UpStudy AI · Accepted Answer

1. For the limit to exist at $$ x=3 $$, the left-hand limit must equal the right-hand limit.

2. The left-hand limit as $$ x \to 3^- $$ is given by:
   $$
     \lim_{x \to 3^-} f(x) = \frac{1}{3} \cdot 3 + c = 1 + c.
   $$

3. The right-hand limit as $$ x \to 3^+ $$ is:
   $$
     \lim_{x \to 3^+} f(x) = -3 + 9 = 6.
   $$

4. Setting the two limits equal:
   $$
     1 + c = 6.
   $$

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

Thus, the value of $$ c $$ that ensures the limit exists is $$ c = 5 $$.
To ensure the limit exists at $$ x=3 $$, set the left-hand limit equal to the right-hand limit. Calculating both: - Left-hand limit: $$ \frac{1}{3} \cdot 3 + c = 1 + c $$ - Right-hand limit: $$ -3 + 9 = 6 $$ Setting them equal: $$ 1 + c = 6 $$, so $$ c = 5 $$. Therefore, $$ c = 5 $$ ensures the limit exists.

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