Answer:
The value of limit = [tex]\frac{13}{8}[/tex]
Step-by-step explanation:
Explanation:-
Given [tex]\lim_{x \to 4} \frac{x^{2}+5 x-36 }{x^{2}-16 }[/tex]
= [tex]\lim_{x \to 4} \frac{x^{2}+9 x-4 x-36 }{(x-4)(x+4) }[/tex]
= [tex]\lim_{x \to 4} \frac{(x(x-4)+9(x-4) }{(x-4)(x+4) }[/tex]
=[tex]\lim_{x \to 4} \frac{((x-4)(x+9)}{(x-4)(x+4) }[/tex]
= [tex]\frac{4+9}{4+4}[/tex]
= [tex]\frac{13}{8}[/tex]
Hence the limit exists and The value of limit = [tex]\frac{13}{8}[/tex]
