2): f ( x ) = x 4 − 27 x 2 {\displaystyle \ f(x)=x^{4}-27x^{2}}
cristianmejia
3): f ( x ) = x 3 − 4 x 2 + 4 x − 1 {\displaystyle \ f(x)=x^{3}-4x^{2}+4x-1}
lim x → − 3 x 2 − x − 12 x + 3 {\displaystyle \lim _{x\to -3}{\frac {x^{2}-x-12}{x+3}}}
= ( − 3 ) 2 − ( − 3 ) − 12 ( − 3 ) + 3 {\displaystyle {\frac {(-3)^{2}-(-3)-12}{(-3)+3}}} = 9 + 3 − 12 0 {\displaystyle {\frac {9+3-12}{0}}} = 0 0 {\displaystyle {\frac {0}{0}}}
= lim x → − 3 ( x + 3 ) ( x − 4 ) x + 3 {\displaystyle \lim _{x\to -3}{\frac {(x+3)(x-4)}{x+3}}} = lim x → − 3 x − 4 = − 7 {\displaystyle \lim _{x\to -3}{x-4}=-7}
lim x → 0 ( 2 − t ) − 2 t {\displaystyle \lim _{x\to 0}{\frac {{\sqrt {(}}2-t)-{\sqrt {2}}}{t}}}
lim x → 0 ( 2 − t ) − 2 t {\displaystyle \lim _{x\to 0}{\frac {{\sqrt {(}}2-t)-{\sqrt {2}}}{t}}} = ( 2 − 0 ) − 2 0 {\displaystyle {\frac {{\sqrt {(}}2-0)-{\sqrt {2}}}{0}}} = 2 − 2 0 = 0 0 {\displaystyle {\frac {{\sqrt {2}}-{\sqrt {2}}}{0}}={\frac {0}{0}}}
lim x → 0 ( 2 − t ) − 2 t {\displaystyle \lim _{x\to 0}{\frac {{\sqrt {(}}2-t)-{\sqrt {2}}}{t}}} = lim x → 0 ( ( 2 − t ) − 2 ) ( ( 2 − t ) + 2 t ( ( 2 − t ) + 2 {\displaystyle \lim _{x\to 0}{\frac {({\sqrt {(}}2-t)-{\sqrt {2}})({\sqrt {(}}2-t)+{\sqrt {2}}}{t({\sqrt {(}}2-t)+{\sqrt {2}}}}} = lim x → 0 ( ( 2 − t ) ) 2 − ( 2 ) 2 t ( ( 2 − t ) + 2 ) {\displaystyle \lim _{x\to 0}{\frac {({\sqrt {(}}2-t))^{2}-({\sqrt {2}})^{2}}{t({\sqrt {(}}2-t)+{\sqrt {2}})}}} = lim x → 0 ( 2 − t ) − 2 ) t ( ( 2 − t ) + 2 ) {\displaystyle \lim _{x\to 0}{\frac {(2-t)-2)}{t({\sqrt {(}}2-t)+{\sqrt {2}})}}} = lim x → 0 − t t ( ( 2 − t ) + 2 ) {\displaystyle \lim _{x\to 0}{\frac {-t}{t({\sqrt {(}}2-t)+{\sqrt {2}})}}} = lim x → 0 − 1 ( 2 − t ) + 2 = − 1 ( 2 − 0 ) + 2 {\displaystyle \lim _{x\to 0}{\frac {-1}{{\sqrt {(}}2-t)+{\sqrt {2}}}}={\frac {-1}{{\sqrt {(}}2-0)+{\sqrt {2}}}}} = − 1 ( 2 ) + 2 {\displaystyle {\frac {-1}{{\sqrt {(}}2)+{\sqrt {2}}}}} = − 1 2 2 {\displaystyle {\frac {-1}{2{\sqrt {2}}}}}
lim x → 0 x ( 1 + 3 x ) − 1 {\displaystyle \lim _{x\to 0}{\frac {x}{{\sqrt {(}}1+3x)-1}}}
lim x → 0 x ( 1 + 3 x ) − 1 {\displaystyle \lim _{x\to 0}{\frac {x}{{\sqrt {(}}1+3x)-1}}} = 0 ( 1 + 3 ( 0 ) ) − 1 {\displaystyle {\frac {0}{{\sqrt {(}}1+3(0))-1}}} = 0 1 − 1 {\displaystyle {\frac {0}{{\sqrt {1}}-1}}} = 0 0 {\displaystyle {\frac {0}{0}}}
lim x → 0 x ( 1 + 3 x ) − 1 {\displaystyle \lim _{x\to 0}{\frac {x}{{\sqrt {(}}1+3x)-1}}} = lim x → 0 x ( ( 1 + 3 x ) + 1 ) ( ( 1 + 3 x ) − 1 ) ( ( 1 + 3 x ) + 1 ) {\displaystyle \lim _{x\to 0}{\frac {x({\sqrt {(}}1+3x)+1)}{({\sqrt {(}}1+3x)-1)({\sqrt {(}}1+3x)+1)}}} = lim x → 0 x ( ( 1 + 3 x ) + 1 ) ( ( 1 + 3 x ) ) 2 ( 1 ) 2 {\displaystyle \lim _{x\to 0}{\frac {x({\sqrt {(}}1+3x)+1)}{({\sqrt {(}}1+3x))^{2}(1)^{2}}}} = lim x → 0 x ( ( 1 + 3 x ) + 1 ) ( 1 + 3 x ) − ( 1 ) {\displaystyle \lim _{x\to 0}{\frac {x({\sqrt {(}}1+3x)+1)}{(1+3x)-(1)}}} = lim x → 0 x ( ( 1 + 3 x ) + 1 ) 3 x ) {\displaystyle \lim _{x\to 0}{\frac {x({\sqrt {(}}1+3x)+1)}{3x)}}} = lim x → 0 ( ( 1 + 3 x ) + 1 ) 3 = ( ( 1 + 3 ( 0 ) ) + 1 ) 3 {\displaystyle \lim _{x\to 0}{\frac {({\sqrt {(}}1+3x)+1)}{3}}={\frac {({\sqrt {(}}1+3(0))+1)}{3}}} = ( 1 + 1 ) 3 = 2 3 {\displaystyle {\frac {({\sqrt {1}}+1)}{3}}={\frac {2}{3}}}
lim x → 1 x 3 − 1 x 2 − 1 {\displaystyle \lim _{x\to 1}{\frac {{\sqrt[{3}]{x}}-1}{{\sqrt[{2}]{x}}-1}}}
lim x → 1 x 3 − 1 x 2 − 1 {\displaystyle \lim _{x\to 1}{\frac {{\sqrt[{3}]{x}}-1}{{\sqrt[{2}]{x}}-1}}} = 1 3 − 1 1 2 − 1 {\displaystyle {\frac {{\sqrt[{3}]{1}}-1}{{\sqrt[{2}]{1}}-1}}} = 1 − 1 1 − 1 {\displaystyle {\frac {1-1}{1-1}}} = 0 0 {\displaystyle {\frac {0}{0}}} Indeterminado
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