Question

Calculo de Limites...

Answer 1

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$\tt a)\\\displaystyle\sf\lim_{x \to -1}\dfrac{x-2}{x^2+4x-3}=\dfrac{-1-1}{(-1)^2+4\cdot(-1)-3}=\dfrac{-2}{1-4-3}=\dfrac{-2}{-6}\\\huge\boxed{\boxed{\boxed{\boxed{\displaystyle\sf\lim_{x \to -1}\dfrac{x-1}{x^2+4x-3}=\dfrac{1}{3}}}}}\checkmark$

$\tt b)\\\displaystyle\sf\lim_{x \to -2}\dfrac{x+2}{x^2-x-6}\\\sf x^2-x-6=x^2+(2-3)x+(2)\cdot(-3)=(x+2)\cdot(x-3)\\\displaystyle\sf\lim_{x \to -2}\dfrac{\diagup\!\!\!\!\!(x+\diagup\!\!\!\!\!2)}{\diagup\!\!\!\!\!(x+\diagup\!\!\!\!\!2)\cdot(x-3)}=\dfrac{1}{-2-3}=-\dfrac{1}{5}\\\huge\boxed{\boxed{\boxed{\boxed{\displaystyle\sf\lim_{x \to -2}\dfrac{x+2}{x^2-x-6}=-\dfrac{1}{5}}}}}\checkmark$

$\tt c)\\\displaystyle\sf\lim_{ h \to 0}\dfrac{(h-5)^2-25}{h}\implies\lim_{h \to 0}\dfrac{[(h-5)-5][(h-5)+5]}{h}\\\displaystyle \lim_{h \to 0}\dfrac{[h-5-5][h-\diagup\!\!\!5+\diagup\!\!\!5]}{h}\implies\lim_{h \to 0}\dfrac{\diagdown\!\!\!\!h\cdot[h-10]}{\diagdown\!\!\!h}\\\displaystyle\lim_{h \to 0}h-10=0-10=-10\\\huge\boxed{\boxed{\boxed{\boxed{\displaystyle\sf\lim_{h \to 0}\dfrac{(h-5)^2-25}{h}=-10}}}}\checkmark$