$$ \begin{array}{ll}\lim _{x ightarrow a^{-}} \frac{x-3}{x-a}= \ \lim _{x ightarrow a^{+}} \frac{x-3}{x-a}= & ext { R B } \ \lim _{x ightarrow a} \frac{x-3}{x-a}=\square\end{array} $$

Question

$$ \begin{array}{ll}\lim _{x ightarrow a^{-}} \frac{x-3}{x-a}= \ \lim _{x ightarrow a^{+}} \frac{x-3}{x-a}= & 	ext { R B } \ \lim _{x ightarrow a} \frac{x-3}{x-a}=\square\end{array} $$

UpStudy AI · Accepted Answer

We will first calculate the left-hand limit $$ \lim_{x \rightarrow a^{-}} \frac{x-3}{x-a} $$.
To find the left-hand limit $$ \lim_{x \rightarrow a^{-}} \frac{x-3}{x-a} $$, we can substitute values approaching $$ a $$ from the left. As $$ x $$ approaches $$ a $$, the denominator $$ x - a $$ approaches $$ 0 $$ and the numerator approaches $$ a - 3 $$. Thus, we can express this limit as:$$
\lim_{x \rightarrow a^{-}} \frac{x-3}{x-a} = \frac{a-3}{0^{-}} = -\infty \text{ (if } a \neq 3\text{)}
$$
Next, we will calculate the right-hand limit $$ \lim_{x \rightarrow a^{+}} \frac{x-3}{x-a} $$.
To find the right-hand limit $$ \lim_{x \rightarrow a^{+}} \frac{x-3}{x-a} $$, we can substitute values approaching $$ a $$ from the right. As $$ x $$ approaches $$ a $$, the denominator $$ x - a $$ approaches $$ 0 $$ and the numerator approaches $$ a - 3 $$. Thus, we can express this limit as:$$
\lim_{x \rightarrow a^{+}} \frac{x-3}{x-a} = \frac{a-3}{0^{+}} = +\infty \text{ (if } a \neq 3\text{)}
$$
Now, we will calculate the overall limit $$ \lim_{x \rightarrow a} \frac{x-3}{x-a} $$. Since the left-hand limit approaches $$ -\infty $$ and the right-hand limit approaches $$ +\infty $$, we can conclude:$$
\lim_{x \rightarrow a} \frac{x-3}{x-a} = \text{undefined (if } a \neq 3\text{)}
$$
The limits are as follows: - $$ \lim_{x \rightarrow a^{-}} \frac{x-3}{x-a} = -\infty $$ - $$ \lim_{x \rightarrow a^{+}} \frac{x-3}{x-a} = +\infty $$ - $$ \lim_{x \rightarrow a} \frac{x-3}{x-a} = \text{undefined (if } a \neq 3\text{)} $$

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