2 If $$ f(x)=\left\{\begin{array}{ll}\frac{x^{2}+6}{a} & , x3\end{array} ight. $$ and $$ \operatorname{Lim}_{x ightarrow 3} f(x) $$ is exist then find the value of $$ a b $$.

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2 If $$ f(x)=\left\{\begin{array}{ll}\frac{x^{2}+6}{a} & , x<3 \ 3 b & , x>3\end{array}ight. $$ and $$ \operatorname{Lim}_{x ightarrow 3} f(x) $$ is exist then find the value of $$ a b $$.

UpStudy AI · Accepted Answer

To determine the value of $$ ab $$ given the function:

$$
f(x) = \begin{cases}
\frac{x^{2} + 6}{a} & \text{if } x < 3 \
3b & \text{if } x > 3
\end{cases}
$$

we need to ensure that the limit of $$ f(x) $$ as $$ x $$ approaches 3 exists. This requires that the left-hand limit (as $$ x $$ approaches 3 from below) and the right-hand limit (as $$ x $$ approaches 3 from above) are equal.

1. **Left-hand limit (as $$ x $$ approaches 3 from below):**

$$
\lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} \frac{x^{2} + 6}{a} = \frac{3^{2} + 6}{a} = \frac{15}{a}
$$

2. **Right-hand limit (as $$ x $$ approaches 3 from above):**

$$
\lim_{x \to 3^+} f(x) = 3b
$$

For the limit to exist, these two expressions must be equal:

$$
\frac{15}{a} = 3b
$$

Solving for $$ ab $$:

$$
15 = 3ab \implies ab = 5
$$

**Answer:** 5
$$ ab = 5 $$

2 If and is exist then find the value of .

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