Question

Stie cineva si ma poate ajuta?va rog frumo? E urgent!

Answer 1

Răspuns:

Explicație pas cu pas:

Prmele 6 sunt de tipul 0/0, se aplica

L'hospital, se deriveaza sus si jos

Ia) lim x->5(1/(2√(x+4)))/(2x) = (1/(2√9))/10 = 10/3

b) lim x->3(4x -3)/(3x^2) = (12-3)/(27) = 9/27 = 1/3

c) lim x->10((cos(x^2 -100)*2x)/(2) = (cos0 * 20)/2 = 1*20/2 = 10

d) lim x ->0(2^x*ln2 +3^x*ln3 +4^x*ln4 +5^x*ln5)/1 =

                  1*ln2 +1*ln3 +1*ln4 +1*ln5 = ln2 +ln3 +ln4 +ln5

e) lim x->0((2007/(1 +2007x))/(2x +9) =

            (2007/1)/9 = 2007/9 = 223

etc.

                 

Answer 2

$I.~a)~ \lim_{x \to 5}\cfrac{\sqrt{x+4}-3 }{x^2-25} \overset{[\frac{0}{0}] }= \lim_{x \to 5} \cfrac{\Big(\sqrt{x+4}-3\Big)' }{\Big(x^2-25\Big)'} =\\\\=\lim_{x \to 5}\cfrac{\Big(\sqrt{x+4}\Big)' -3'}{\Big(x^2\Big)'-(25)'} =\lim_{x \to 5}\cfrac{\cfrac{1}{2\sqrt{x+4}}\cdot (x+4)' }{2x^{2-1}-0} =\\\\=\lim_{x \to 5}\cfrac{\cfrac{1}{2\sqrt{x+4} } }{2x}=\lim_{x \to 5}\Bigg(\cfrac{1}{2\sqrt{x+4} } \cdot \cfrac{1}{2x}\Bigg)=\lim_{x \to 5}\cfrac{1}{4x\sqrt{x+4} }=$

$=\cfrac{1}{4\cdot5\sqrt{5+4} }=\cfrac{1}{20\sqrt{9} }=\cfrac{1}{20\cdot3}= \cfrac{1}{60}$

$b)~ \lim_{x \to 3} \cfrac{2x^2-3x-9}{x^3-27}\overset{[\frac{0}{0}] }=\lim_{x \to 3}\cfrac{\Big(2x^2-3x-9\Big)'}{\Big(x^3-27\Big)'}=\\\\=\lim_{x \to 3}\cfrac{2 \cdot 2x-3 \cdot 1-0}{3x^2-0} =\lim_{x \to 3}\cfrac{4x-3}{3x^2} =\cfrac{4 \cdot 3-3}{3 \cdot 3^2} =\\\\=\cfrac{12-3}{3 \cdot 9} =\cfrac{9}{27}= \cfrac{1}{3}$

$c)~ \lim_{x \to 10} \cfrac{sin\Big(x^2-100\Big)}{2x-20} \overset{[\frac{0}{0}] }=\lim_{x \to 10}\cfrac{\Big(sin\Big(x^2-100\Big)\Big)'}{(2x-20)' } =\\\\=\lim_{x \to 10}\cfrac{cos\Big(x^2-100\Big)\cdot\Big(x^2-100\Big)'}{2-0} =\lim_{x \to 10}\cfrac{cos\Big(x^2-100\Big)\cdot2x}{2} =\\\\=\cfrac{cos\Big(10^2-100\Big)\cdot 2 \cdot 10}{2} =\cfrac{cos(100-100)\cdot 20}{2} =\cfrac{cos~0 \cdot 20}{2} =\cfrac{1 \cdot 20}{2} =10$

$d)~\lim_{x \to 0}\cfrac{2^x+3^x+4^x+5^x-4}{x} \overset{[\frac{0}{0}] }=\lim_{x \to 0}\cfrac{\Big(2^x+3^x+4^x+5^x-4\Big)'}{x'}=\\\\=\lim_{x \to 0}\cfrac{2^xln2+3^xln3+4^xln4+5^xln5-0}{1} =\\\\=\lim_{x \to 0}\Big(2^xln2+3^xln3+4^xln4+5^xln5\Big)=\\\\=2^0ln2+3^0ln3+4^0ln4+5^0ln5= 1\cdot ln2+1\cdot ln3+1\cdot ln4+1 \cdot ln5=\\\\=ln2+ln3+ln4+ln5=ln(2\cdot3\cdot4\cdot5)=ln(120)$

$e)~\lim_{x \to 0}\cfrac{ln(1+2007x)}{x^2+9x} \overset{[\frac{0}{0} ]}=\lim_{x \to 0}\cfrac{(ln(1+2007x))'}{\Big(x^2+9x\Big)'}=\\\\=\lim_{x \to 0}\cfrac{\cfrac{1}{1+2007x}\cdot(1+2007x)' }{2x+9} =\lim_{x \to 0}\cfrac{\cfrac{1}{1+2007x}\cdot 2007 }{2x+9} =\\\\=\lim_{x \to 0}\cfrac{2007}{(1+2007x)(2x+9)} =\cfrac{2007}{(1+2007\cdot0)(2\cdot0+9)} =\\\\=\cfrac{2007}{(1+0)(0+9)}=\cfrac{2007}{1\cdot9}=\cfrac{2007}{9} =223$

$f)~\lim_{x \to 1}\cfrac{\sqrt{2x^2+x+1}-\sqrt{x^2+x+2} }{\sqrt{2x+7}-\sqrt{x+8} }}\overset{[\frac{0}{0} ]}=\\\\=\lim_{x \to 1}\cfrac{\Big(\sqrt{2x^2+x+1}-\sqrt{x^2+x+2}\Big) ' }{\Big(\sqrt{2x+7}-\sqrt{x+8}\Big)' } =\\\\=\lim_{x \to 1}\cfrac{\cfrac{1}{2\sqrt{2x^2+x+1} }\cdot\Big(2x^2+x+1\Big)'-\cfrac{1}{2\sqrt{x^2+x+2} }\cdot\Big(x^2+x+2\Big)' }{\cfrac{1}{2\sqrt{2x+7} }\cdot(2x+7)'-\cfrac{1}{2\sqrt{x+8} }\cdot(x+8)' } =$

$=\lim_{x \to 1}\cfrac{\cfrac{4x+1}{2\sqrt{2x^2+x+1} }-\cfrac{2x+1}{2\sqrt{x^2+x+2} } }{\cfrac{2}{2\sqrt{2x+7} }-\cfrac{1}{2\sqrt{x+8} } } =\\\\=\cfrac{\cfrac{4 \cdot 1+1}{2\sqrt{2\cdot1^2+1+1} }-\cfrac{2\cdot1+1}{2\sqrt{1^2+1+2} } }{\cfrac{2}{2\sqrt{2\cdot1+7} }-\cfrac{1}{2\sqrt{1+8} } }=\cfrac{\cfrac{5}{2\sqrt{4} }-\cfrac{3}{2\sqrt{4} } }{\cfrac{2}{2\sqrt{9} }-\cfrac{1}{2\sqrt{9} } } =\cfrac{\cfrac{5-3}{2 \cdot 2} }{\cfrac{2-1}{2 \cdot 3} }=\cfrac{\cfrac{2}{4} }{\cfrac{1}{6} }=$

$=\cfrac{2}{4} \cdot 6=\cfrac{12}{4} =3$

$II.~a)~ \lim_{x \to \infty} \cfrac{8x^2+x+5}{4x^2+x+1} \overset{[\frac{\infty}{\infty} ]}=\lim_{x \to \infty}\cfrac{\Big(8x^2+x+5\Big)'}{\Big(4x^2+x+1\Big)'}=\\\\=\lim_{x \to \infty}\cfrac{8 \cdot 2x+1+0}{4 \cdot 2x+1+0} = \lim_{x \to \infty}\cfrac{16x+1}{8x+1} =\lim_{x \to \infty}\cfrac{16(x+1)}{8(x+1)} =2$

$b)~\lim_{x \to \infty}\cfrac{5x^2+x+5}{2x+4} =\lim_{x \to \infty}\overset{[\frac{\infty}{\infty} ]}=\lim_{x \to \infty}\cfrac{\Big(5x^2+x+5\Big)'}{(2x+4)} =\\\\=\lim_{x \to \infty}\cfrac{5 \cdot 2x+1+0}{2 \cdot 1+0} =\lim_{x \to \infty}\cfrac{10x+1}2} =\cfrac{10 \cdot \infty +1}{2} =\cfrac{\infty}{2} =\infty$

$c)~\lim_{x \to \infty}\cfrac{10x+8}{x^2+x+9}\overset{[\frac{\infty}{\infty} ]}=\lim_{x \to \infty} \cfrac{(10x+8)'}{\Big(x^2+x+9\Big)'} =\lim_{x \to \infty} \cfrac{10 \cdot 1 + 0}{2x+1+0} =\\\\=\lim_{x \to \infty}\cfrac{10}{2x+1} =\cfrac{10}{2\cdot\infty+1} =\cfrac{10}{\infty} =0$

$d)~\lim_{x \to \infty}\dfrac{\sqrt{4x^2+3x+2}+\sqrt{x^2+x+1} }{x+7} \overset{[\frac{\infty}{\infty} ]}=\\\\= \lim_{x \to \infty}\cfrac{\sqrt{x^2\Bigg(4+\cfrac{3}{x} +\cfrac{2}{x^2} }\Bigg)+\sqrt{x^2\Bigg(1+\cfrac{1}{x} +\cfrac{1}{x^2}} \Bigg) }{x\Bigg(1+\cfrac{7}{x} \Bigg)} =\\\\= \lim_{x \to \infty} \cfrac{\sqrt{x^2} \Bigg(\sqrt{4+\cfrac{3}{x}+\cfrac{2}{x^2}}\Bigg) +\sqrt{x^2}\Bigg(\sqrt{1+\cfrac{1}{x}+\cfrac{1}{x^2} }\Bigg) }{x\Bigg(1+\cfrac{7}{x} \Bigg)} =$

$=\cfrac{\sqrt{4+0+0}+\sqrt{1+0+0} }{1+0} =\cfrac{\sqrt{4} +\sqrt{1} }{1} =2+1=3$