Matemática, perguntado por oliveira0225196, 4 meses atrás

urgente
Prove cada proposição utilizando a definição , de limite.​

Anexos:

Soluções para a tarefa

Respondido por VitiableIndonesia
1

\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{ \color{green} \boxed{Olá\:tudo\:bem?}}}}}}}

~\huge\mid{\boxed{\bf{\blue{Matem\acute{a}tica}}}\mid}

A) Verdadeiro.

\bold{\underset{x\rightarrow2}{\lim}~}( {x}^{2}  - 4x + 5) = 1

\bold{\underset{x\rightarrow2}{\lim}~}( {x}^{2}  - 4x) + \bold{\underset{x\rightarrow2}{\lim}~}(5)

\bold{\underset{x\rightarrow2}{\lim}~}( {x}^{2}) -  \bold{\underset{x\rightarrow2}{\lim}~}(4x) + \bold{\underset{x\rightarrow2}{\lim}~}(5)

\bold{\underset{x\rightarrow2}{\lim}~}( {x}^{2} ) - \bold{\underset{x\rightarrow2}{\lim}~}(4x) + 5

(\bold{\underset{x\rightarrow2}{\lim}~}(x) {)}^{2}  - \bold{\underset{x\rightarrow2}{\lim}~}(4x) + 5

(\bold{\underset{x\rightarrow2}{\lim}~}(x) {)}^{2}  - 4 \times \bold{\underset{x\rightarrow2}{\lim}~}(x) + 5

 {2}^{2}  - 4 \times \bold{\underset{x\rightarrow2}{\lim}~}(x) + 5

 {2}^{2} - 4 \times 2 + 5

4 - 4 \times 2 + 5

4 - 8 + 5

\boxed{ \color{green} \boxed{{ 1 }}}

B) Verdadeiro.

\bold{\underset{x\rightarrow - 2}{\lim}~}( {x}^{2}  - 1)

\bold{\underset{x\rightarrow - 2}{\lim}~}( {x}^{2} ) - \bold{\underset{x\rightarrow - 2}{\lim}~}(1)

(\bold{\underset{x\rightarrow - 2}{\lim}~}(x) {)}^{2}  - \bold{\underset{x\rightarrow - 2}{\lim}~}(1)

(\bold{\underset{x\rightarrow - 2}{\lim}~}(x) {)}^{2}  - 1

( - 2 {)}^{2}  - 1

4 - 1

\boxed{ \color{green} \boxed{{ 3 }}}

C) Verdadeiro.

\bold{\underset{x\rightarrow3}{\lim}~}( \frac{ {x}^{2}  + x - 12}{x - 3}

\bold{\underset{x\rightarrow3}{\lim}~}( {x}^{2} + x - 12) \\  \bold{\underset{x\rightarrow3}{\lim}~}(x - 3) \:  \:  \:  \:  \:  \:  \:  \:  \:  \:

 {3}^{2}  + 3 - 12 \\ \bold{\underset{x\rightarrow3}{\lim}~}(x - 3)

0 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \\ \bold{\underset{x\rightarrow3}{\lim}~}(x - 3)

0  \:  \:  \:  \:  \:  \: \\ 3 - 3

0 \\ 0

\bold{\underset{x\rightarrow3}{\lim}~}( \frac{ {x}^{2}  + x - 12}{x - 3} )

\bold{\underset{x\rightarrow3}{\lim}~}( \frac{ {x}^{2}  + 4x - 3x - 12}{x - 3} )

\bold{\underset{x\rightarrow3}{\lim}~}( \frac{x \times (x  + 4) - 3x - 12}{x - 3} )

\bold{\underset{x\rightarrow3}{\lim}~}( \frac{x \times (x + 4) - 3(x + 4)}{x - 3} )

\bold{\underset{x\rightarrow3}{\lim}~}( \frac{(x + 4) \times (x - 3)}{x - 3} )

\bold{\underset{x\rightarrow3}{\lim}~}( \frac{(x + 4) \times \cancel{ (x - 3) }}{\cancel{ x -  3 }})

\bold{\underset{x\rightarrow3}{\lim}~}(x + 4)

\bold{\underset{x\rightarrow3}{\lim}~}(x) + \bold{\underset{x\rightarrow3}{\lim}~}(4)

3 + \bold{\underset{x\rightarrow3}{\lim}~}(4)

3 + 4

\boxed{ \color{green} \boxed{{ 7 }}}

D) Verdadeiro.

\bold{\underset{x\rightarrow9}{\lim}~}( \sqrt{9 - x} )

 \sqrt{\bold{\underset{x\rightarrow9}{\lim}~}(9 - x)}

 \sqrt{\bold{\underset{x\rightarrow9}{\lim}~}(9) - \bold{\underset{x\rightarrow9}{\lim}~}(x)}

 \sqrt{9 - \bold{\underset{x\rightarrow9}{\lim}~}(x)}

 \sqrt{9 - 9}

 \sqrt{0}

\boxed{ \color{green} \boxed{{ 0 }}}

{\huge\boxed { {\bf{E}}}\boxed { \red {\bf{a}}} \boxed { \blue {\bf{s}}} \boxed { \gray{\bf{y}}} \boxed { \red {\bf{}}} \boxed { \orange {\bf{M}}} \boxed {\bf{a}}}{\huge\boxed { {\bf{t}}}\boxed { \red {\bf{h}}}}

{\boxed{ \color{blue} \boxed{ 26 |02|22  }}}{\boxed{ \color{blue} \boxed{Espero \:  ter  \: ajudado \: ☆}}}


oliveira0225196: muito obrigada pela ajuda
VitiableIndonesia: de nada
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