Plz solve this problem of the chapter " LIMITS"

1.	Plz solve this problem of the chapter " LIMITS"
Answer» QUESTION 1: $\displaystyle \lim_{x \to 3}( {x}^{<klux>2</klux>} + \sqrt{3 - x} )$ ANSWER 1: 9 EXPLANATION 1: $\displaystyle \lim_{x \to 3}( {x}^{2} + \sqrt{3 - x} )\\ \\ \\{3}^{2} + \sqrt{3 - 3} \\ \\ \\ 9\\ \\ \\ \\{\large\bf{\displaystyle \lim_{x \to 3}( {x}^{2} + \sqrt{3 - x} ) = 9}}$ QUESTION 2: $\displaystyle \lim_{x \to 2} \left( \dfrac{1}{3 + {e}^{ \dfrac{1}{x - 2} } } \right )$ ANSWER 2: 0 FORMULA: $\boxed{ \large{{{e}^{ \infty } } =\infty = \frac{1}{0} }}$ $\boxed{ \large{ \dfrac{1}{\dfrac{1}{0}} = 0 }}$ EXPLANATION 2: $\displaystyle \lim_{x \to 2} \left( \dfrac{1}{3 + {e}^{ \dfrac{1}{2- 2} } } \right) \\ \\ \\ \displaystyle \lim_{x \to 2} \left( \dfrac{1}{3 + {e}^{ \infty } } \right) \\ \\ \\ \dfrac{1}{3 + \dfrac{1}{0} } \\ \\ \\ \dfrac{1}{ \dfrac{0 + 1}{0} } \\ \\ \\ \dfrac{1}{ \dfrac{1}{0} } \\ \\ \\ \bf{ \large{ \displaystyle \lim_{x \to 2} \left( \dfrac{1}{3 + {e}^{ \dfrac{1}{2- 2} } } \right) = 0 }}$

Answer»

QUESTION 1:

$\displaystyle \lim_{x \to 3}( {x}^{<klux>2</klux>} + \sqrt{3 - x} )$

EXPLANATION 1:

$\displaystyle \lim_{x \to 3}( {x}^{2} + \sqrt{3 - x} )\\ \\ \\{3}^{2} + \sqrt{3 - 3} \\ \\ \\ 9\\ \\ \\ \\{\large\bf{\displaystyle \lim_{x \to 3}( {x}^{2} + \sqrt{3 - x} ) = 9}}$

QUESTION 2:

$\displaystyle \lim_{x \to 2} \left( \dfrac{1}{3 + {e}^{ \dfrac{1}{x - 2} } } \right )$

ANSWER 2:

FORMULA:

$\boxed{ \large{{{e}^{ \infty } } =\infty = \frac{1}{0} }}$

$\boxed{ \large{ \dfrac{1}{\dfrac{1}{0}} = 0 }}$

EXPLANATION 2:

$\displaystyle \lim_{x \to 2} \left( \dfrac{1}{3 + {e}^{ \dfrac{1}{2- 2} } } \right) \\ \\ \\ \displaystyle \lim_{x \to 2} \left( \dfrac{1}{3 + {e}^{ \infty } } \right) \\ \\ \\ \dfrac{1}{3 + \dfrac{1}{0} } \\ \\ \\ \dfrac{1}{ \dfrac{0 + 1}{0} } \\ \\ \\ \dfrac{1}{ \dfrac{1}{0} } \\ \\ \\ \bf{ \large{ \displaystyle \lim_{x \to 2} \left( \dfrac{1}{3 + {e}^{ \dfrac{1}{2- 2} } } \right) = 0 }}$