To evaluate the limit [tex]\(\lim_{x \to \pi / 2} \frac{\sin^2 x + 5 \sin x - 6}{\sin x - 1}\)[/tex], we can proceed as follows:
1. Substitute [tex]\(\sin x\)[/tex] with a new variable:
Let [tex]\( y = \sin x \)[/tex]. As [tex]\( x \)[/tex] approaches [tex]\( \pi/2 \)[/tex], [tex]\( \sin x \)[/tex] approaches 1. Therefore, we will analyze how the function behaves as [tex]\( y \)[/tex] approaches 1.
The limit then becomes:
[tex]\[
\lim_{y \to 1} \frac{y^2 + 5y - 6}{y - 1}
\][/tex]
2. Simplify the numerator:
Factor the quadratic expression in the numerator [tex]\( y^2 + 5y - 6 \)[/tex]:
[tex]\[
y^2 + 5y - 6 = (y + 6)(y - 1)
\][/tex]
This allows us to rewrite the limit as:
[tex]\[
\lim_{y \to 1} \frac{(y + 6)(y - 1)}{y - 1}
\][/tex]
3. Cancel the common factor:
Notice that [tex]\( y - 1 \)[/tex] appears in both the numerator and the denominator, so we can cancel it out:
[tex]\[
\lim_{y \to 1} \frac{(y + 6)\cancel{(y - 1)}}{\cancel{(y - 1)}} = \lim_{y \to 1} (y + 6)
\][/tex]
4. Evaluate the limit:
As [tex]\( y \)[/tex] approaches 1, we simply substitute [tex]\( y \)[/tex] with 1 in the remaining expression:
[tex]\[
(y + 6) \bigg|_{y = 1} = 1 + 6 = 7
\][/tex]
Therefore, the value of the limit is:
[tex]\[
\lim_{x \to \pi/2} \frac{\sin^2 x + 5 \sin x - 6}{\sin x - 1} = 7
\][/tex]
