Limits of Functions: Problems with Solutions

Solution:
$\lim_{x\to3}\frac{x^2-3x}{x-3}=\lim_{x\to3}\frac{x(x-3)}{x-3}=\lim_{x\to3}x=3$

Solution:
$\lim_{x\to-1}(2x^3-4x^2+5)=\lim_{x\to-1}(2x^3)-\lim_{x\to-1}(4x^2)+\lim_{x\to-1}5=-2-4+5=-1$

Solution:
$\lim_{x\to0}\frac{6x^3-5x^2+4x}{2x}=\lim_{x\to0}\frac{x(6x^2-5x+4)}{2x}=\lim_{x\to0}\frac{6x^2-5x+4}{2}=$

$\frac{\lim_{x\to0}(6x^2)-\lim_{x\to0}(5x)+\lim_{x\to0}4}{\lim_{x\to0}2}=\frac{0-0+4}{2}=2$

Solution:
$\lim_{x\to2}\frac{5x-10}{x^2-x-2}=\lim_{x\to2}\frac{5(x-2)}{(x+1)(x-2)}=\frac{\lim_{x\to2}5}{\lim_{x\to2}(x+1)}=\frac{\lim_{x\to2}5}{\lim_{x\to2}(x)+\lim_{x\to2}(1)}=\frac{5}{2+1}=\frac{5}{3}$

Solution:
$\lim_{x\to0}\frac{5x^2}{3x^3+2x^2}=\lim_{x\to0}\frac{5x^2}{x^2(3x+2}=\lim_{x\to0}\frac{5}{3x+2}=\frac{\lim_{x\to0}5}{\lim_{x\to0}(3x)+\lim_{x\to0}2}=\frac{\lim_{x\to0}5}{\lim_{x\to0}3\cdot \lim_{x\to0}x+\lim_{x\to0}2}=\frac{5}{3\cdot0+2}=\frac{5}{2}$

Solution:
$\lim_{x\to1}\left(\frac{x}{2x-2}-\frac{2-x}{x^2-1}\right)=\lim_{x\to1}\left(\frac{x}{2(x-1)}-\frac{2-x}{(x-1)(x+1)}\right)=$

$\lim_{x\to1}\left(\frac{x(x+1)}{2(x-1)(x+1)}-\frac{(2-x)2}{2(x-1)(x+1)}\right)=\lim_{x\to1}\frac{x^2+x-4+2x}{2(x-1)(x+1)}=\lim_{x\to1}\frac{x^2+3x-4}{2(x-1)(x+1)}=\lim_{x\to1}\frac{(x+4)(x-1)}{2(x-1)(x+1)}=\lim_{x\to1}\frac{x+4}{2(x+1)}=$

$\frac{\lim_{x\to1}x+\lim_{x\to1}4}{\lim_{x\to1}2\cdot(\lim_{x\to1}x+\lim_{x\to1}1)}=\frac{1+4}{2(1+1)}=\frac{5}{4}$

Solution:
$\lim_{x\to \infin}\frac{6x^3-x+5}{8x^3+3x-10}=\lim_{x\to \infin}\frac{x^3(6-\frac{1}{x^2}+\frac{5}{x^3})}{x^3(8+\frac{3}{x^2}-\frac{10}{x^3})}=\lim_{x\to \infin}\frac{6-\frac{1}{x^2}+\frac{5}{x^3}}{8+\frac{3}{x^2}-\frac{10}{x^3}}=$

$\frac{\lim_{x\to \infin}6-\lim_{x\to \infin}\frac{1}{x^2}+\lim_{x\to \infin}\frac{5}{x^3}}{\lim_{x\to \infin}8+\lim_{x\to \infin}\frac{3}{x^2}-\lim_{x\to \infin}\frac{10}{x^3}}=$

$\frac{\lim_{x\to \infin}6-\lim_{x\to \infin}\frac{1}{x}\cdot\lim_{x\to \infin}\frac{1}{x}+\lim_{x\to \infin}5\cdot\lim_{x\to \infin}\frac{1}{x}\cdot\lim_{x\to \infin}\frac{1}{x}\cdot\lim_{x\to \infin}\frac{1}{x}}{\lim_{x\to \infin}8+\lim_{x\to \infin}3\cdot\lim_{x\to \infin}\frac{1}{x}\cdot\lim_{x\to \infin}\frac{1}{x}-\lim_{x\to \infin}10\cdot\lim_{x\to \infin}\frac{1}{x}\cdot\lim_{x\to \infin}\frac{1}{x}\cdot\lim_{x\to \infin}\frac{1}{x}}=$

$\frac{6-\lim_{x\to \infin}\frac{1}{x}\cdot\lim_{x\to \infin}\frac{1}{x}+5\cdot\lim_{x\to \infin}\frac{1}{x}\cdot\lim_{x\to \infin}\frac{1}{x}\cdot\lim_{x\to \infin}\frac{1}{x}}{8+3\cdot\lim_{x\to \infin}\frac{1}{x}\cdot\lim_{x\to \infin}\frac{1}{x}-10\cdot\lim_{x\to \infin}\frac{1}{x}\cdot\lim_{x\to \infin}\frac{1}{x}\cdot\lim_{x\to \infin}\frac{1}{x}}=$

$\frac{6-0\times0+5\times0\times0\times0}{8+3\times0\times0-10\times0\times0\times0}=\frac{6}{8}=\frac{3}{4}$

Solution:
$\lim_{x\to 0^{+}}\frac{x^2-1}{x^2+x}=\lim_{x\to 0^{+}}\frac{(x-1)(x+1)}{x(x+1)}=\lim_{x\to 0^{+}}\frac{x-1}{x}=$

$\lim_{x\to 0^{+}}\frac{x(1-\frac{1}{x})}{x}=\lim_{x\to 0^{+}}1-\frac{1}{x}=\lim_{x\to 0^{+}}1-\lim_{x\to 0^{+}}\frac{1}{x}=1-\infin=-\infin$

Solution:
$\lim_{x\to\infin}\frac{x+2}{x^2+4x+5}=\lim_{x\to\infin}\frac{x(1+\frac{2}{x})}{x^2(1+\frac{4}{x}+\frac{5}{x^2})}=$

$\lim_{x\to\infin}\frac{1+\frac{2}{x}}{x(1+\frac{4}{x}+\frac{5}{x^2})}=\lim_{x\to\infin}\frac{1}{x}\cdot \lim_{x\to\infin}\frac{1+\frac{2}{x}}{1+\frac{4}{x}+\frac{5}{x^2}}=$ $0\cdot\lim_{x\to\infin}\frac{1+\frac{2}{x}}{1+\frac{4}{x}+\frac{5}{x^2}}=0$

Solution:
$\lim_{x\to\infin}\frac{8x^4+5x^3}{4x^3+2x^2-x}=\lim_{x\to\infin}\frac{x^4(8+\frac{5}{x})}{x^3(4+\frac{2}{x}-\frac{1}{x^2})}=\lim_{x\to\infin}\frac{x(8+\frac{5}{x})}{4+\frac{2}{x}-\frac{1}{x^2}}=$

$\frac{\lim_{x\to\infin}x\lim_{x\to\infin}(8+\frac{5}{x})}{\lim_{x\to\infin}4+\lim_{x\to\infin}\frac{2}{x}-\lim_{x\to\infin}\frac{1}{x^2}}=\frac{\lim_{x\to\infin}x(\lim_{x\to\infin}8+\lim_{x\to\infin}\frac{5}{x})}{4+\lim_{x\to\infin}(2\cdot\frac{1}{x})-\lim_{x\to\infin}(\frac{1}{x}\cdot\frac{1}{x})}=$

$\frac{\lim_{x\to\infin}x(\lim_{x\to\infin}8+\lim_{x\to\infin}5\cdot\lim_{x\to\infin}\frac{1}{x})}{4+\lim_{x\to\infin}2\cdot\lim_{x\to\infin}\frac{1}{x}-\lim_{x\to\infin}\frac{1}{x}\cdot\lim_{x\to\infin}\frac{1}{x}}=\frac{\infin\cdot(8+5\cdot0)}{4+2\cdot0-0\cdot0}=\frac{\infin\cdot8}{4}=2\cdot\infin=\infin$

Solution:
$\lim_{x\to\infin}\frac{3x^2-10x}{3x-7x^2} = \lim_{x\to\infin}\frac{x^2(3-\frac{10}{x})}{x^2(\frac{3}{x}-7)} = \lim_{x\to\infin}\frac{3-\frac{10}{x}}{\frac{3}{x}-7} =$

$\frac{\lim_{x\to\infin}3-\lim_{x\to\infin}\frac{10}{x}}{\lim_{x\to\infin}\frac{3}{x}- \lim_{x\to\infin}7}= \frac{3-\lim_{x\to\infin}(10\cdot\frac{1}{x})}{\lim_{x\to\infin}(3\cdot\frac{1}{x})- 7}= \frac{3-\lim_{x\to\infin}10\cdot\lim_{x\to\infin}\frac{1}{x}}{\lim_{x\to\infin}3\cdot\lim_{x\to\infin}\frac{1}{x}- 7}=$

$\frac{3-10\cdot0}{3\cdot0-7}=-\frac{3}{7}$

Solution:
$\lim_{x\to5}\frac{x-5}{\sqrt{x-1}-2}=\lim_{x\to5}\frac{x-5}{\sqrt{x-1}-2}\cdot\frac{\sqrt{x-1}+2}{\sqrt{x-1}+2}=$

$\lim_{x\to5}\frac{(x-5)(\sqrt{x-1}+2)}{x-1-4}=\lim_{x\to5}\frac{(x-5)(\sqrt{x-1}+2)}{x-5}=\lim_{x\to5}(\sqrt{x-1}+2)=$

$\lim_{x\to5}\sqrt{x-1}+\lim_{x\to5}2=\sqrt{4}+2=4$

Solution:
$\lim_{x\to1}\frac{x^2-2x+1}{x^3-x^2-x+1}=\lim_{x\to1}\frac{(x-1)^2}{x^2(x-1)-(x-1)}=\lim_{x\to1}\frac{(x-1)^2}{(x-1)(x^2-1)}=$

$\lim_{x\to1}\frac{x-1}{x^2-1}=\lim_{x\to1}\frac{x-1}{(x-1)(x+1)}=\lim_{x\to1}\frac{1}{x+1}=\frac{1}{1+1}=\frac{1}{2}$

Solution:
$\lim_{x\to2}\frac{10-3x-x^2}{x^4-9x^2+20}=\lim_{x\to2}\frac{-(x^2+3x-10}{x^4-9x^2+20}=\lim_{x\to2}\frac{-(x-2)(x+5)}{(x^2-5)(x-2)(x+2)}=$

$\lim_{x\to2}\frac{-(x+5)}{(x^2-5)(x+2)}=\lim_{x\to2}\frac{-(2+5)}{(2^2-5)(2+2)}=\frac{-7}{-4}=\frac{7}{4}$

Solution:
$\lim_{x\to7}\frac{\sqrt{x-3}-2}{x-7}=\lim_{x\to7}\frac{\sqrt{x-3}-2}{x-7}\cdot\frac{\sqrt{x-3}+2}{\sqrt{x-3}+2}=$

$\lim_{x\to7}\frac{(\sqrt{x-3})^2-2^2}{(x-7)(\sqrt{x-3}+2)}=\lim_{x\to7}\frac{x-3-4}{(x-7)(\sqrt{x-3}+2)}=\lim_{x\to7}\frac{x-7}{(x-7)(\sqrt{x-3}+2)}=$

$\lim_{x\to7}\frac{1}{\sqrt{x-3}+2}=\frac{1}{\sqrt{7-3}+2}=\frac{1}{2+2}=\frac{1}{4}$

Solution:
$\lim_{x\to0}\frac{x}{\sin x}=\lim_{x\to0}\frac{\frac{x}{x}}{\frac{\sin x}{x}}=\lim_{x\to0}\frac{1}{\frac{\sin x}{x}}=$

$\frac{\lim_{x\to0}1}{\lim_{x\to0}\frac{\sin x}{x}}=\frac{1}{1}=1$

Solution:
$\lim_{x\to0}\frac{\sin 5x}{x}=\lim_{x\to0}\frac{5\sin 5x}{5x}=\lim_{x\to0}5\cdot\lim_{x\to0}\frac{\sin 5x}{5x}=5\cdot1=5$

Solution:
$\lim_{x\to 0}\frac{3\sin 3x}{x}=\lim_{x\to 0}\frac{3\cdot3\sin 3x}{3x}=\lim_{x\to 0}\left(9\frac{\sin 3x}{3x}\right)=\lim_{x\to 0}9\lim_{x\to 0}\frac{\sin 3x}{3x}=9\cdot1=9$

Solution:
$\lim_{x\to0}\frac{\sin^3\frac{x}{2}}{x^3}=\lim_{x\to0}\left(\frac{\sin\frac{x}{2}}{x}\right)^3=\lim_{x\to0}\left(\frac{1}{2}\cdot\frac{\sin\frac{x}{2}}{\frac{x}{2}}\right)^3=$

$\lim_{x\to0}\left(\frac{1}{2}\right)^3\lim_{x\to0}\left(\frac{\sin\frac{x}{2}}{\frac{x}{2}}\right)^3=\frac{1}{8}\cdot1=\frac{1}{8}$

Solution:
$\lim_{x\to0}\frac{1-\cos x}{x^2}=\lim_{x\to0}\frac{2\sin^2\frac{x}{2}}{x^2}=\lim_{x\to0}2\left(\frac{\sin\frac{x}{2}}{x}\right)^2=$

$\lim_{x\to0}2\left(\frac{\sin\frac{x}{2}}{\frac{2x}{2}}\right)^2=\lim_{x\to0}2\left(\frac{1}{2}\cdot\frac{\sin\frac{x}{2}}{\frac{x}{2}}\right)^2=\lim_{x\to0}2\cdot\frac{1}{4}\cdot\left(\frac{\sin\frac{x}{2}}{\frac{x}{2}}\right)^2=$

$\lim_{x\to0}\frac{1}{2}\cdot\left(\frac{\sin\frac{x}{2}}{\frac{x}{2}}\right)^2=\lim_{x\to0}\frac{1}{2}\cdot\left(\lim_{x\to0}\frac{\sin\frac{x}{2}}{\frac{x}{2}}\right)^2=\frac{1}{2}\cdot1=\frac{1}{2}$

Solution:
$\lim_{x\to1}\frac{(x^2-6x+5)\sin(x-1)}{(x-1)^2}=\lim_{x\to1}\frac{(x-1)(x-5)\sin(x-1)}{(x-1)^2}=\lim_{x\to1}\frac{(x-5)\sin(x-1)}{x-1}=$

$\lim_{x\to1}(x-5)\cdot\lim_{x\to1}\frac{\sin(x-1)}{x-1}=(1-5)\cdot1=-4$

Solution:
$\lim_{x\to0}\frac{\cos x-\cos3x}{x^2}=\lim_{x\to0}\frac{-(\cos3x-\cos x)}{x^2}=\lim_{x\to0}\frac{-(-2\sin\frac{3x+x}{2}\sin\frac{3x-x}{2})}{x^2}=$

$\lim_{x\to0}\frac{2\sin2x\sin x}{x\cdot x}=\lim_{x\to0}\frac{2\sin 2x\sin x}{\frac{1}{2}\cdot2\cdot x\cdot x}=$

$\lim_{x\to0}\frac{2}{\frac{1}{2}}\cdot\lim_{x\to0}\frac{\sin2x}{2x}\cdot\lim_{x\to0}\frac{\sin x}{x}=4\cdot1\cdot1=4$

Solution:
$\lim_{x\to0}\frac{\sin^2x}{1-\cos x}=\lim_{x\to0}\frac{1-\cos^2x}{1-\cos x}=\lim_{x\to0}\frac{(1-\cos x)(1+\cos x)}{1-\cos x}=$

$\lim_{x\to0}(1+\cos x) = \lim_{x\to0}1 + \lim_{x\to0}\cos x= 1+1=2$

Solution:
$\lim_{x\to0}\frac{\sin 5x-\sin 3x}{\sin x}=\lim_{x\to0}\frac{2\cos\frac{5x+3x}{2}\cdot\sin\frac{5x-3x}{2}}{\sin x}=$

$\lim_{x\to0}\frac{2\cos\frac{8x}{2}\cdot\sin\frac{2x}{2}}{\sin x}=\lim_{x\to0}\frac{2\cos4x\cdot\sin x}{\sin x}=$

$\lim_{x\to0}(2\cos4x)=\lim_{x\to0}2\cdot\lim_{x\to0}(\cos4x)=2\cdot1=2$

Solution:
$\lim_{x\to2}\frac{\sqrt{7+x}-\sqrt{11-x}}{\sqrt{x+2}-\sqrt{3x-2}}=$

$\lim_{x\to2}\frac{\sqrt{7+x}-\sqrt{11-x}}{\sqrt{x+2}-\sqrt{3x-2}}\cdot\frac{\sqrt{7+x}+\sqrt{11-x}}{\sqrt{7+x}+\sqrt{11-x}}\cdot\frac{\sqrt{x+2}+\sqrt{3x-2}}{\sqrt{x+2}+\sqrt{3x-2}}=$

$\lim_{x\to2}\frac{(\sqrt{7+x}-\sqrt{11-x})(\sqrt{7+x}+\sqrt{11-x})(\sqrt{x+2}+\sqrt{3x-2})}{(\sqrt{x+2}-\sqrt{3x-2})(\sqrt{x+2}+\sqrt{3x-2})(\sqrt{7+x}+\sqrt{11-x})}=$

$\lim_{x\to2}\frac{(7+x-11+x)(\sqrt{x+2}+\sqrt{3x-2})}{(x+2-3x+2)(\sqrt{x+2}+\sqrt{3x-2})(\sqrt{7+x}+\sqrt{11-x})}=$

$\lim_{x\to2}\frac{(2x-4)(\sqrt{x+2}+\sqrt{3x-2})}{(4-2x)(\sqrt{7+x}+\sqrt{11-x})}=\lim_{x\to2}\frac{(2x-4)(\sqrt{x+2}+\sqrt{3x-2})}{-(2x-4)(\sqrt{7+x}+\sqrt{11-x})}=$

$-\lim_{x\to2}\frac{\sqrt{x+2}+\sqrt{3x-2}}{\sqrt{7+x}+\sqrt{11-x}}=-\frac{\sqrt{2+2}+\sqrt{6-2}}{\sqrt{7+2}+\sqrt{11-2}}=-\frac{2+2}{3+3}=-\frac{2}{3}$