Answer:
A) 1.25
Step-by-step explanation:
Use L'Hôpital's Rule and differentiate both top and bottom expressions:
[tex]\lim_{x \to 3} \frac{x^{2}-x-6 }{x^{2}-2x-3}[/tex]
[tex]\lim_{x \to 3} \frac{2x-1-0 }{2x-2-0}[/tex]
[tex]\lim_{x \to 3} \frac{2x-1}{2x-2}[/tex]
Substitute x=3 into the expression:
[tex]\frac{2(3)-1}{2(3)-2}[/tex]
[tex]\frac{6-1}{6-2}[/tex]
[tex]\frac{5}{4}[/tex]
Therefore, [tex]\lim_{x \to 3} \frac{x^{2}-x-6 }{x^{2}-2x-3}=\frac{5}{4}=1.25[/tex], making the correct answer A.
