Question

Am nevoie de ajutor la limite.
Exercițiile 4 si 5, punctele a si b de la amandoua. Am atașat poza.
Dau puncte
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Answer 1

$\displaystyle 4.~a)~ \lim_{x \to 1} \frac{\sqrt{x} -1}{\sqrt[3]{x}-1 } \overset{[\frac{0}{0} ]}=\lim_{x \to 1} \frac{\Big(\sqrt{x} -1\Big)'}{\Big(\sqrt[3]{x}-1 \Big)'} =\lim_{x \to 1} \frac{\Big(\sqrt{x} \Big)'-1'}{\Big(\sqrt[3]{x}\Big)'-1' } =$

$\displaystyle =\lim_{x \to 1} \frac{\cfrac{1}{2\sqrt{x} }-0 }{\Big(x^{\frac{1}{3} }\Big)'-0} =\lim_{x \to 1} \frac{\cfrac{1}{2\sqrt{x} } }{\cfrac{1}{3} \cdot x^{\frac{1}{3}-1 }} =\lim_{x \to 1} \frac{\cfrac{1}{2\sqrt{x} } }{\cfrac{1}{3} \cdot x^{\frac{1-3}{3} } } =\lim_{x \to 1} \frac{\cfrac{1}{2\sqrt{x} } }{\cfrac{1}{3} \cdot x^{-\frac{2}{3} }} =$

$\displaystyle =\lim_{x \to 1}\frac{\cfrac{1}{2\sqrt{x} } }{\cfrac{1}{3} \cdot \cfrac{1}{x^{\frac{2}{3} }} } =\frac{\cfrac{1}{2\sqrt{1} } }{\cfrac{1}{3}\cdot \cfrac{1}{1^{\frac{2}{3} }} } =\frac{\cfrac{1}{2} }{\cfrac{1}{3} } =\frac{1}{2} \cdot 3=\frac{3}{2}$

$\displaystyle b)~\lim_{x \to 7} \frac{2-\sqrt{x-3} }{x^2-49} \overset{[\frac{0}{0} ]}=\lim_{x \to 7}\frac{\Big(2-\sqrt{x-3}\Big)' }{\Big(x^2-49\Big)'} =\lim_{x \to 7}\frac{2'-\Big(\sqrt{x-3}\Big)' }{\Big(x^2\Big)'-(49)'} =$

$\displaystyle =\lim_{x \to 7}\frac{0-\cfrac{1}{2\sqrt{x-3} } \cdot (x-3)'}{2x-0} =\lim_{x \to 7}\frac{-\cfrac{1}{2\sqrt{x-3} }\cdot (x'-3') }{2x} =$

$\displaystyle =\lim_{x \to 7}\frac{-\cfrac{1}{2\sqrt{x-3} }\cdot (1-0) }{2x} =\lim_{x \to 7}\frac{-\cfrac{1}{2\sqrt{x-3} } \cdot 1}{2x} =\lim_{x \to 7}\frac{-\cfrac{1}{2\sqrt{x-3} } }{2x} =$

$\displaystyle =\frac{-\cfrac{1}{2\sqrt{7-3} } }{2\cdot 7}=\frac{-\cfrac{1}{2\sqrt{4} } }{14} =\frac{-\cfrac{1}{2\cdot2} }{14} =\frac{-\cfrac{1}{4} }{14} =-\frac{1}{4} \cdot \frac{1}{14}=- \frac{1}{56}$

$\displaystyle 5.~a)~\lim_{x \to 2}\frac{x^2-7x+10}{x^2-8x+12} \overset{[\frac{0}{0} ]}=\lim_{x \to 2}\frac{\Big(x^2-7x+10\Big)'}{\Big(x^2-8x+12\Big)'} =\lim_{x \to 2}\frac{\Big(x^2\Big)'-7 x'+(10)'}{\Big(x^2\Big)'-8 x'+(12)'}=$

$\displaystyle =\lim_{x \to 2}\frac{2x-7 \cdot 1+0}{2x-8 \cdot 1+0} =\lim_{x \to 2}\frac{2x-7}{2x-8} =\frac{2\cdot 2-7}{2\cdot 2-8} =\frac{4-7}{4-8} =\frac{-3}{-4} =\frac{3}{4}$

$\displaystyle b)~ \lim_{x \to +\infty}\frac{2^x+3}{2^x-3} \overset{[\frac{\infty}{\infty}] }=\lim_{x \to +\infty}\frac{\Big(2^x+3\Big)'}{\Big(2^x-3\Big)'} =\lim_{x \to +\infty}\frac{\Big(2^x\Big)'+3'}{\Big(2^x\Big)'-3'} =$

$\displaystyle =\lim_{x \to +\infty}\frac{2^xln2+0}{2^xln2-0} =\lim_{x \to +\infty}\frac{2^xln2}{2^xln2} =\lim_{x \to +\infty}1=1$