Question

Evaluate the following limits​

Answer 1

Answer:

y²+ 216

this is the answer

Answer 2

Answer:

Step-by-step explanation:

$\lim_{y \to -3} \frac{y^{5}+243 }{y^{3}+27 }$

Since, the given limit is of 0/0 form. using L - Hospital rule we get

$\lim_{y \to -3} \frac{5y^{4}+0 }{3y^{2}+0}$

= $\lim_{y \to -3} \frac{5y^{4} }{3y^{2}}$

$=\frac{5(-3)^{4} }{3(-3)^{2} }$

$=15$