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Answer 1

Answer:

• Refer to the attachment ✔️

Answer 2

$\large\rm{ \lim\limits_{x\to3} \Bigg [ \dfrac{x^2-9}{x-3} \Bigg ] }$

since it is in 0/0 form apply L'hopital's rule

$\large\rm$

$\large\rm{ = \lim\limits_{x\to3} \dfrac{x^2-9}{x-3}}$

$\large\rm$

$\large\rm{ = \lim\limits_{x\to3} \dfrac{(x^2-9)'}{(x-3)'}}$

$\large\rm$

$\large\rm{ = \lim\limits_{x\to3} \dfrac{2x}{1}}$

$\large\rm$

$\large\rm{ = \dfrac{2 \times 3}{1} = \dfrac{6}{1} = 6}$