Matematică
paulilie32
16

am si eu nevoie de ajutor urgent cine imi raspunde ii dau coroana va rog ex C15

+1
(2) Răspunsuri
mmika

a) [latex] \lim_{n \to \11} \frac{x^{2}+2x-3 }{ x^{2} -5x+4} = \frac{0}{0} -caz-de- nedeterminare =\ \textgreater \ aplici-L'h[/latex] [latex] \frac{ (x^{2} +2x-3)'}{( x^{2} -5x+4)'} [/latex]  =>[latex] \lim_{x \to \11} \frac{2x+2}{2x-5} = \frac{4}{-3} [/latex] Asa faci la toate 0/0 sau infinit/infinit.Derivezi ce e sus pe derivata celui de jos pana ti se reduc si scapi de nedeterminare

Simonatopolog

[latex]\displaystyle \mathtt{C15.~a)~ \lim_{x \to 1} \frac{x^2+2x-3}{x^2-5x+4} \overset{\underset{\mathrm{l'H}}{}}{=} \lim_{x \to 1} \frac{\left(x^2+2x-3\right)'}{\left(x^2-5x+4\right)'} =}\\ \\ \mathtt{= \lim_{x \to 1} \frac{\left(x^2\right)'+(2x)'-3'}{\left(x^2\right)'-(5x)'+4'} = \lim_{x \to 1} \frac{2x+2}{2x-5}= \frac{2\cdot1+2}{2\cdot1-5}= \frac{2+2}{2-5}=- \frac{4}{3} }[/latex] [latex]\displaystyle \mathtt{b)~ \lim_{x \to \infty} \frac{ln~x}{x}\overset{\underset{\mathrm{l'H}}{}}{=} \lim_{x \to \infty} \frac{(ln~x)'}{x'}= \lim_{x \to \infty} \frac{ \frac{1}{x} }{1}= \lim_{x \to \infty} \frac{1}{x}= \frac{1}{\infty}=0}[/latex] [latex]\displaystyle \mathtt{c)~ \lim_{x \to \infty} \frac{e^x}{x^2} \overset{\underset{\mathrm{l'H}}{}}{=} \frac{\left(e^x\right)'}{\left(x^2\right)'} = \frac{e^x}{2x} \overset{\underset{\mathrm{l'H}}{}}{=} \lim_{x \to \infty}\frac{\left(e^x\right)'}{(2x)'}= \lim_{x \to \infty} \frac{e^x}{2x'}= \lim_{x \to \infty} \frac{e^x}{2 \cdot 1} =}\\ \\ \mathtt{= \lim_{x \to \infty} \frac{e^x}{2} = \frac{1}{2} \lim_{x \to \infty}e^x= \frac{1}{2}\cdot \infty =\infty }[/latex] [latex]\displaystyle \mathtt{d)~ \lim_{x \to 0} \frac{e^x-x-1}{x^2}\overset{\underset{\mathrm{l'H}}{}}{=} \lim_{x \to 0} \frac{\left(e^x-x-1\right)'}{\left(x^2\right)'}= \lim_{x \to 0} \frac{\left(e^x\right)'-x'-1'}{2x}= }\\ \\ \mathtt{=\lim_{x \to 0} \frac{e^x-1}{2x}\overset{\underset{\mathrm{l'H}}{}}{=} \lim_{x \to 0} \frac{\left(e^x-1\right)'}{(2x)'}= \lim_{x \to 0} \frac{\left(e^x\right)'-1'}{2}=\lim_{x \to 0} \frac{e^x}{2} = \frac{e^ 0}{2} = \frac{1}{2} }[/latex] [latex]\displaystyle \mathtt{e)~\lim_{x \to 1} \frac{ln~x}{x-1}\overset{\underset{\mathrm{l'H}}{}}{=} \lim_{x \to 1} \frac{(ln~x)'}{(x-1)'}=\lim_{x \to 1} \frac{ \frac{1}{x} }{x'-1'}= \lim_{x \to 1} \frac{ \frac{1}{x} }{1}= \lim_{x \to 1} \frac{1}{x} = \frac{1}{1} =1 }[/latex] [latex]\displaystyle \mathtt{f)~\lim_{x \to 0} \frac{2e^x-x-2}{sin~x}\overset{\underset{\mathrm{l'H}}{}}{=}\lim_{x \to 0} \frac{\left(2e^x-x-2\right)'}{(sin~x)'}= \lim_{x \to 0} \frac{\left(2e^x\right)'-x'-2}{cos~x}= }\\ \\ \mathtt{=\lim_{x \to 0} \frac{2e^x-1}{cos~x}= \frac{2e^0-1}{cos~0}= \frac{2 \cdot 1-1}{1}=2-1=1 }[/latex] [latex]\displaystyle \mathtt{g)~\lim_{x \to 0} \frac{1-cos~x}{x^2}\overset{\underset{\mathrm{l'H}}{}}{=}\lim_{x \to 0} \frac{(1-cos~x)'}{\left(x^2\right)'}= \lim_{x \to 0} \frac{1'-(cos~x)'}{2x} =}\\ \\ \mathtt{=\lim_{x \to 0} \frac{sin~x}{2x}\overset{\underset{\mathrm{l'H}}{}}{=} \lim_{x \to 0} \frac{(sin~x)'}{(2x)'}=\lim_{x \to 0} \frac{cos~x}{2} = \frac{cos~0}{2} = \frac{1}{2} }[/latex] [latex]\displaystyle \mathtt{h)~\lim_{x \to 0} \frac{2sin~x+cos~x-1}{x^2-x}\overset{\underset{\mathrm{l'H}}{}}{=}\lim_{x \to 0} \frac{(2sin~x+cos~x-1)'}{\left(x^2-x\right)'}=}\\ \\ \mathtt{=\lim_{x \to 0} \frac{(2sin~x)'+(cos~x)'-1'}{\left(x^2\right)'-x'} =\lim_{x \to 0} \frac{2(sin~x)'-sin~x}{2x-1} =}\\ \\ \mathtt{=\lim_{x \to 0} \frac{2cos~x-sin~x}{2x-1} = \frac{2cos~0-sin~0}{2 \cdot 0-1}= \frac{2 \cdot 1-0}{0-1} =-2 }[/latex]

Adaugă răspuns