Question

Ma ajutati va rog? E urgent

Answer 1

$\displaystyle \mathtt{1.~a)~ \lim_{x \to 1} \frac{x^2+2x-3}{3x^2-5x+2}=\lim_{x \to 1} \frac{(x-1)(x+3)}{(x-1)(3x-2)}=\lim_{x \to 1} \frac{x+3}{3x-2} = }\\ \\ \mathtt{= \frac{1+3}{3\cdot1-2}= \frac{4}{3-2}= 4}$

$\displaystyle \mathtt{b)~\lim_{x \to 2} \frac{x^3-2x^2-4x+8}{x^3-4x^2+4x}=\lim_{x \to 2} \frac{x^2(x-2)-4(x-2)}{x\left(x^2-4x+4\right)}=}\\ \\ \mathtt{=\lim_{x \to 2} \frac{\left(x^2-4\right)(x-2)}{x(x-2)^2}= \lim_{x \to 2} \frac{x^2-4}{x(x-2)}= \lim_{x \to 2} \frac{(x-2)(x+2)}{x(x-2)}= }\\ \\ \mathtt{=\lim_{x \to 2} \frac{x+2}{x} = \frac{2+2}{2}= \frac{4}{2} =2 }$

$\displaystyle \mathtt{c)~\lim_{x \to 1} \frac{ln(2x-1)}{ln(3x-2)} \overset{\underset{\mathrm{l'H}}{}}{=} \lim_{x \to 1}\frac{[ln(2x-1)]'}{[ln(3x-2)]'}=\lim_{x \to 1} \frac{ \frac{(2x-1)'}{2x-1} }{ \frac{(3x-2)'}{3x-2} }=\lim_{x \to 1} \frac{ \frac{(2x)'-1'}{2x-1} }{ \frac{(3x)'-2'}{3x-2} } =}\\ \\ \mathtt{=\lim_{x \to 1} \frac{ \frac{2}{2x-1} }{ \frac{3}{3x-2} } =\lim_{x \to 1} \frac{2(3x-2)}{3(2x-1)} =\lim_{x \to 1} \frac{6x-4}{6x-3} = \frac{6\cdot1-4}{6\cdot1-3} = \frac{6-4}{6-3}= \frac{2}{3} }$

$\displaystyle \mathtt{d)~ \lim_{x \to 0} \frac{sin~x+x^2}{x^3+2x} \overset{\underset{\mathrm{l'H}}{}}{=} \lim_{x \to 0} \frac{\left(sin~x+x^2\right)'}{\left(x^3+2x\right)'}=\lim_{x \to 0} \frac{(sin~x)'+\left(x^2\right)'}{\left(x^3\right)'+(2x)'}= }\\ \\ \mathtt{=\lim_{x \to 0} \frac{cos~x+2x}{3x^2+2} = \frac{cos~0+2 \cdot 0}{3 \cdot 0^2+2}= \frac{1+0}{0+2} = \frac{1}{2} }$

$\displaystyle \mathtt{e)~\lim_{x \to 1} \frac{ln~x}{x^2-3x+2} \overset{\underset{\mathrm{l'H}}{}}{=}\lim_{x \to 1} \frac{(ln~x)'}{\left(x^2-3x+2\right)'}=\lim_{x \to 1} \frac{ \frac{1}{x} }{\left(x^2\right)'-(3x)'+2'}= }\\ \\ \mathtt{=\lim_{x \to 1} \frac{ \frac{1}{x} }{2x-3} =\lim_{x \to 1} \frac{1}{x(2x-3)}=\lim_{x \to 1} \frac{1}{2x^2-3x}= \frac{1}{2\cdot1^2-3\cdot1} =}\\ \\ \mathtt{= \frac{1}{2-3} = \frac{1}{-1}=-1 }$

$\displaystyle \mathtt{f)~\lim_{x \to 1}\frac{\sqrt{x}-1}{\sqrt[\mathtt3]{\mathtt x} -1}\overset{\underset{\mathrm{l'H}}{}}{=}\lim_{x \to1}\frac{\left(\sqrt{x}-1\right)'}{\left(\sqrt[\mathtt3]{\mathtt x}-1\right)'}=\lim_{x\to 1}\frac{\left(\sqrt{x} \right)'-1'}{\left(\sqrt[\mathtt3]{\mathtt x}\right)'-1'}=\lim_{x\to1}\frac{\frac{1}{2\sqrt{x}}}{\frac{1}{3x^{\frac{2}{3}}}}=}$

$\displaystyle \mathtt{=\lim_{x\to 1}\frac{3x^{\frac{2}{3}}}{2x^{\frac{1}{2}}}=\lim_{x\to1}\frac{3x^{\frac{1}{6}}}{2}=\frac{3\cdot1^{\frac{1}{6}}}{2}=\frac{3}{2}}$

$\displaystyle \mathtt{2.~a)~ \lim_{x \to \infty} \frac{x^3-x^2+2x-ln~3}{-x^2+2x-5} =\lim_{x \to \infty} \frac{x-1+ \frac{2}{x}- \frac{ln~3}{x^2} }{-1+ \frac{2}{x}- \frac{5}{x^2} }= }\\\\\mathtt{= \frac{\infty-1+ \frac{2}{\infty} - \frac{ln~3}{\infty^2} }{-1+ \frac{2}{\infty}- \frac{5}{\infty^2}}=\frac{\infty-1+0-0}{-1+0-0}= \frac{\infty}{-1}=-\infty}$

$\displaystyle \mathtt{b)~\lim_{x \to \infty} \frac{x^2+x+5}{2x+1}=\lim_{x \to \infty} \frac{x+1+ \frac{5}{x} }{2+ \frac{1}{x}}= \frac{\infty+1+ \frac{5}{\infty} }{2+ \frac{1}{\infty}}= \frac{\infty+1+0}{2+0}=\infty}$

$\displaystyle \mathtt{c)~\lim_{x \to \infty} \frac{ln~x-x^3+2}{x^3+x+1} =\lim_{x \to \infty} \frac{ \frac{ln~x}{x^3} -1+ \frac{2}{x^3} }{1+ \frac{1}{x^2}+ \frac{1}{x^3} }=\frac{ \frac{ln~\infty}{\infty^3}-1+ \frac{2}{\infty^3}}{1+ \frac{1}{\infty^2} + \frac{1}{\infty^3}}=}\\ \\ \mathtt{=\frac{0-1+0}{1+0+0}= \frac{-1}{1}=-1 }}$

$\displaystyle \mathtt{3.~a)~ \lim_{x \to 2} \frac{x^3-8}{x^7-128}\overset{\underset{\mathrm{l'H}}{}}{=} \lim_{x \to 2} \frac{\left(x^3-8\right)'}{\left(x^7-128\right)'}=\lim_{x \to 2} \frac{\left(x^3\right)'-8'}{\left(x^7\right)'-128'}=}\\ \\ \mathtt{=\lim_{x \to 2} \frac{3x^2}{7x^6} = \frac{3 \cdot 2^2}{7 \cdot 2^6}= \frac{3}{7 \cdot 2^4}= \frac{3}{7 \cdot 16}=\frac{3}{112} }$

$\displaystyle \mathtt{b)~\lim_{x \to 0} \frac{e^{3x}-1}{e^{4x}-1}\overset{\underset{\mathrm{l'H}}{}}{=}\lim_{x \to 0} \frac{\left(e^{3x}-1\right)'}{\left(e^{4x}-1\right)'}=\lim_{x \to 0} \frac{\left(e^{3x}\right)'-1'}{\left(e^{4x}\right)'-1'}=\lim_{x \to 0} \frac{3e^{3x}}{4e^{4x}} =} \\ \\ \mathtt{= \frac{3e^{3 \cdot 0}}{4e^{4 \cdot 0}}= \frac{3e^0}{4e^0}= \frac{3 \cdot 1}{4 \cdot 1} = \frac{3}{4} }$

$\displaystyle \mathtt{c)~\lim_{x \to 1} \frac{2^x-x-1}{x-1} \overset{\underset{\mathrm{l'H}}{}}{=}\lim_{x \to 1} \frac{\left(2^x-x-1\right)'}{(x-1)'}=\lim_{x \to 1} \frac{\left(2^x\right)'-x'-1'}{x'-1'}=}\\ \\ \mathtt{=\lim_{x \to 1} \frac{2^xln~2-1}{1} =\lim_{x \to 1}\left(2^xln~2-1\right)=2^1ln~2-1=2ln~2-1}$