Question

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Answer 1

 

$\displaystyle\bf\\1)\\\\a)\\\lim_{x \to 0}(-2x^2+7x+6)=-2\cdot0^2+7\cdot0+6=0+0+6=\boxed{\bf6}\\\\b)\\\lim_{x\to\infty}(1-3x+7x^2)\\1-3x+7x^2~~sau~~7x^2-3x+1\\este~o~functie~de~gradul~2~in~care~coeficientul~lui~x^2~este~pozitiv.\\Graficul~acestei~functii~este~o~parabola~cu~ramurile~insus,\\care~tind~spre~infinit~atuncii~cand~x~merge~spre~infinit.\\\\\implies~\lim_{x\to\infty}(1-3x+7x^2)=\boxed{\bf +\infty}\\\\c)\\\lim_{x\to1}\frac{x-5}{2x+3}=\frac{1-5}{2\cdot1+3}=\boxed{\bf\frac{-4}{5}}$

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$\displaystyle\bf\\d)\\\lim_{x \to 0}\frac{x^2-5x+8}{x+4}=\frac{0^2-5\cdot0+8}{0+4}=\frac{0-0+8}{0+4}=\frac{8}{4}=\boxed{\bf2}\\\\e)\\\lim_{x\to\infty}5^x=\infty~~(exponentiala~creste~mai~repede~decat~puterea)\\\\f)\\\lim_{x \to 0}\frac{x^3-2x^2}{3x^2+x}=\lim_{x \to 0}\frac{x(x^2-2x)}{x(3x+1)}=\lim_{x \to 0}\frac{x^2-2x}{3x+1}=\frac{0-0}{0+1}=\frac{0}{1}=\boxed{\bf0}\\\\g)\\\lim_{x\to 2}\frac{x^2-2x}{x^2-4}=\frac{x(x-2)}{(x+2)(x-2)}=\frac{x}{x+2}=\frac{2}{2+2}=\frac{2}{4}=\boxed{\bf\frac{1}{2}}$