To solve the limit [tex]\(\lim_{x \rightarrow 1} \frac{1 - x^2}{\sin (\pi x)}\)[/tex], let's proceed step-by-step.
1. Direct Substitution:
First, we try to directly substitute [tex]\(x = 1\)[/tex] into the function:
[tex]\[
\frac{1 - 1^2}{\sin (\pi \cdot 1)} = \frac{1 - 1}{\sin (\pi)} = \frac{0}{0}
\][/tex]
Since this results in the indeterminate form [tex]\(\frac{0}{0}\)[/tex], we need to apply more advanced techniques to evaluate the limit.
2. Simplification and Factoring:
Rewrite the numerator and consider the Taylor series expansion or factoring if applicable. Note that [tex]\(1 - x^2\)[/tex] is a difference of squares:
[tex]\[
1 - x^2 = (1 - x)(1 + x)
\][/tex]
So the function becomes:
[tex]\[
\frac{(1 - x)(1 + x)}{\sin (\pi x)}
\][/tex]
3. L'Hôpital's Rule:
Since we still have an indeterminate form, we can apply L'Hôpital's Rule, which states that for limits of the form [tex]\(\frac{0}{0}\)[/tex] or [tex]\(\frac{\infty}{\infty}\)[/tex], we can take derivatives of the numerator and the denominator:
[tex]\[
\lim_{x \to 1} \frac{1 - x^2}{\sin (\pi x)} = \lim_{x \to 1} \frac{d}{dx} (1 - x^2) / \frac{d}{dx} (\sin (\pi x))
\][/tex]
Differentiating the numerator and the denominator, we get:
[tex]\[
\text{Numerator's derivative: } \frac{d}{dx} (1 - x^2) = -2x
\][/tex]
[tex]\[
\text{Denominator's derivative: } \frac{d}{dx} (\sin (\pi x)) = \pi \cos (\pi x)
\][/tex]
Now, the limit becomes:
[tex]\[
\lim_{x \to 1} \frac{-2x}{\pi \cos (\pi x)}
\][/tex]
4. Direct Substitution Again:
Substitute [tex]\(x = 1\)[/tex] into the simplified derivative limit:
[tex]\[
\frac{-2 \cdot 1}{\pi \cos (\pi \cdot 1)} = \frac{-2}{\pi \cos (\pi)}
\][/tex]
[tex]\[
\cos (\pi) = -1
\][/tex]
So,
[tex]\[
\frac{-2}{\pi \cdot (-1)} = \frac{-2}{-\pi} = \frac{2}{\pi}
\][/tex]
Thus, the limit is:
[tex]\[
\lim_{x \rightarrow 1} \frac{1 - x^2}{\sin (\pi x)} = \frac{2}{\pi}
\][/tex]
This is the final solution.
