Certainly! Let's find the limit of the expression [tex]\(\frac{2x - 3}{x + 5}\)[/tex] as [tex]\(x\)[/tex] approaches 2.
1. Substitute [tex]\(x = 2\)[/tex] into the expression:
[tex]\[
\frac{2(2) - 3}{2 + 5} = \frac{4 - 3}{7} = \frac{1}{7}
\][/tex]
2. Verify there is no indeterminate form:
To ensure we do not get an indeterminate form [tex]\( \frac{0}{0} \)[/tex] or [tex]\( \frac{\infty}{\infty} \)[/tex], substitute [tex]\(x = 2\)[/tex] into both the numerator and the denominator separately:
- Numerator: [tex]\(2x - 3\)[/tex] when [tex]\(x = 2\)[/tex]
[tex]\[
2(2) - 3 = 4 - 3 = 1
\][/tex]
- Denominator: [tex]\(x + 5\)[/tex] when [tex]\(x = 2\)[/tex]
[tex]\[
2 + 5 = 7
\][/tex]
Since both the numerator and the denominator yield finite, non-zero values, this confirms there is no indeterminate form.
3. Conclude the limit:
Since substituting [tex]\(x = 2\)[/tex] directly into the expression gives us finite values:
[tex]\[
\lim_{x \rightarrow 2} \frac{2x - 3}{x + 5} = \frac{1}{7}
\][/tex]
Therefore, the limit is:
[tex]\[
\boxed{\frac{1}{7}}
\][/tex]
