To evaluate the limit [tex]\(\lim_{x \to 1} \frac{x^2 + 2x - 3}{x^2 + 3x - 4}\)[/tex] using L'Hôpital's Rule, we follow these steps:
1. Check the form of the limit:
First, substitute [tex]\( x = 1 \)[/tex] into the function:
[tex]\[
f(x) = x^2 + 2x - 3 \quad \text{and} \quad g(x) = x^2 + 3x - 4.
\][/tex]
[tex]\[
f(1) = 1^2 + 2(1) - 3 = 1 + 2 - 3 = 0,
\][/tex]
[tex]\[
g(1) = 1^2 + 3(1) - 4 = 1 + 3 - 4 = 0.
\][/tex]
Since both the numerator and denominator evaluate to 0 at [tex]\( x = 1 \)[/tex], this is an indeterminate form [tex]\( \frac{0}{0} \)[/tex]. This suggests that L'Hôpital's Rule can be applied.
2. Differentiate numerator and denominator:
Differentiate the numerator [tex]\( f(x) = x^2 + 2x - 3 \)[/tex]:
[tex]\[
f'(x) = \frac{d}{dx}(x^2 + 2x - 3) = 2x + 2.
\][/tex]
Differentiate the denominator [tex]\( g(x) = x^2 + 3x - 4 \)[/tex]:
[tex]\[
g'(x) = \frac{d}{dx}(x^2 + 3x - 4) = 2x + 3.
\][/tex]
3. Apply L'Hôpital's Rule:
Substitute [tex]\( x = 1 \)[/tex] in the derivatives:
[tex]\[
\lim_{x \to 1} \frac{f(x)}{g(x)} = \lim_{x \to 1} \frac{x^2 + 2x - 3}{x^2 + 3x - 4} = \lim_{x \to 1} \frac{f'(x)}{g'(x)} = \lim_{x \to 1} \frac{2x + 2}{2x + 3}.
\][/tex]
Now, substitute [tex]\( x = 1 \)[/tex] into the simplified expression:
[tex]\[
\frac{2(1) + 2}{2(1) + 3} = \frac{2 + 2}{2 + 3} = \frac{4}{5}.
\][/tex]
4. Conclusion:
The evaluated limit is:
[tex]\[
\lim_{x \to 1} \frac{x^2 + 2x - 3}{x^2 + 3x - 4} = \frac{4}{5}.
\][/tex]
Hence, the limit [tex]\(\lim_{x \to 1} \frac{x^2 + 2x - 3}{x^2 + 3x - 4}\)[/tex] evaluates to [tex]\(\frac{4}{5}\)[/tex].
