Soal dan Solusi Limit Fungsi Aljabar

\(\begin{align}
\lim_{x\to -3} 3x+5 &= 3(-3)+5\\
&=-9+5\\
&=-4
\end{align}\)

\(\begin{align}
\lim_{x\to 1} \frac{2x^2-x-1}{3x^2-x-2}&=\lim_{x\to 1} \frac{(2x+1)(x-1)}{(3x+2)(x-1)}\\
&=\lim_{x\to 1}\frac{2x+1}{3x+2}\\
&=\frac{2(1)+1}{3(1)+2}\\
&=\frac{3}{5}\\
\end{align}\)

\(\begin{align}
\lim_{x\to 4} \frac{x^2-2x-8}{x-4}&=\lim_{x\to 4} \frac{(x-4)(x+2)}{x-4}\\
&=\lim_{x\to 4} {(x+2)}\\
&=(4+2)\\
&=6\\
\end{align}\)

\(\begin{align}
\lim_{x\to 1} \frac{2x^2-2}{x-1}&=\lim_{x\to 1} \frac{2(x^2-1)}{x-1}\\
&=\lim_{x\to 1} \frac{2(x-1)(x+1)}{x-1}\\
&=\lim_{x\to 1} 2(x+1)\\
&=2(1+1)\\
&=2(2)\\
&=4\\
\end{align}\)

\(\begin{align}
\lim_{x\to 2} \frac{x^2-5x+6}{x^2-3x+3}&=\frac{2^2-5(2)+6}{2^2-3(2)+3}\\
&=\frac{4-10+6}{4-6+3}\\
&=\frac{0}{1}\\
&=0
\end{align}\)

\(\begin{align}
\lim_{x\to 3} \frac{2x^2-5x-3}{x-3}&=\lim_{x\to 3} \frac{(2x+1)(x-3)}{x-3}\\
&=\lim_{x\to 3} 2x+1\\
&=2(3)+1\\
&=7\\
\end{align}\)

\(\begin{align}
\lim_{x\to -5} \frac{x^2+3x-10}{x+5}&=\lim_{x\to -5} \frac{(x+5)(x-2)}{x+5}\\
&=\lim_{x\to -5} x-2\\
&=-5-2\\
&=-7\\
\end{align}\)

\(\begin{align}
\lim_{x\to 2} \frac{x^4-16}{x-2}&=\lim_{x\to 2} \frac{(x^2-4)(x^2+4)}{x-2}\\
&=\lim_{x\to 2} \frac{(x-2)(x+2)(x^2+4)}{x-2}\\
&=\lim_{x\to 2}(x+2)(x^2+4)\\
&=(2+2)(2^2+4)\\
&=4(4+4)\\
&=4(8)\\
&=32\\
\end{align}\)

\(\begin{align}
\lim_{x\to 2} \frac{x^2-3x+2}{x^2-2x}&=\lim_{x\to 2} \frac{(x-2)(x-1)}{x(x-2)}\\
&=\lim_{x\to 2} \frac{x-1}{x}\\
&=\frac{2-1}{2}\\
&=\frac{1}{2}\\
\end{align}\)

\(\begin{align}
\lim_{x\to 0} \frac{x^3-2x^2+x}{2x^3+x^2-2x}&=\lim_{x\to 0} \frac{x(x^2-2x+1)}{x(2x^2+x-2)}\\
&=\lim_{x\to 0} \frac{x^2-2x+1}{x(x^2+x-2)}\\
&=\lim_{x\to 0} \frac{(x-1)(x-1)}{(x+2)(x-1)}\\
&=\lim_{x\to 0} \frac{x-1}{x+2}\\
&=\frac {0-1}{0+2}\\
&=\frac{-1}{2}\\
\end{align}\)

\(\begin{align}
\lim_{x\to 1} \frac{x^3-5x+4}{x^3-1}&=\lim_{x\to 1} \frac{(x-1)(x^2+x-4)}{(x-1)(x^2+x+1)}\\
&=\lim_{x\to 1} \frac{(x^2+x-4)}{(x^2+x+1)}\\
&=\frac{1^2+1-4}{1^2+1+1}\\
&=\frac{1+1-4}{1+1+1}\\
&=\frac{-2}{3}\\
&=-\frac{2}{3}\\
\end{align}\)

\(\begin{align}
\lim_{x\to -2} \frac{3x+6}{x^3+8}&=\lim_{x\to -2} \frac{3(x+2)}{x^3+2^3}\\
&=\lim_{x\to -2} \frac{3(x+2)}{(x+2)(x^2-2x+4)}\\
&=\lim_{x\to -2} \frac{3}{(x^2-2x+4)}\\
&=\frac{3}{(-2)^2-2(-2)+4)}\\
&=\frac{3}{4+4+4}\\
&=\frac{3}{12}\\
&=\frac{1}{4}\\
\end{align}\)

\(\begin{align}
\lim_{x\to 36} \frac{x-36}{\sqrt{x}-6}&=\lim_{x\to 36} \frac{(\sqrt x)^2-6^2}{\sqrt{x}-6}\\
&=\lim_{x\to 36} \frac{(\sqrt x+6)(\sqrt x-6)}{\sqrt{x}-6}\\
&=\lim_{x\to 36} (\sqrt x+6)\\
&=(\sqrt {36}+6)\\
&=(6+6)\\
&=12
\end{align}\)

\(\begin{align}
\lim_{x\to 1} \left(\frac{1}{x^2-1}-\frac{2}{x^4-1}\right)&=\lim_{x\to 1} \left(\frac{1(x^4-1)-2(x^2-1)}{(x^4-1)(x^2-1)}\right)\\
&=\lim_{x\to 1} \left(\frac{x^4-1-2x^2+2}{(x^4-1)(x^2-1)}\right)\\
&=\lim_{x\to 1} \left(\frac{x^4-2x^2+1}{(x^2-1)(x^2+1)(x^2-1)}\right)\\
&=\lim_{x\to 1} \left(\frac{(x^2-1)(x^2-1)}{(x^2-1)(x^2+1)(x^2-1)}\right)\\
&=\lim_{x\to 1} \frac{1}{x^2+1}\\
&=\frac{1}{(1^2+1)}\\
&=\frac{1}{2}\\
\end{align}\)

\(\begin{align}
\lim_{x\to 2} \left(\frac{6-x}{x^2-4}-\frac{1}{x-2}\right)&=\lim_{x\to 2} \left(\frac{(6-x)-1(x-2)}{x^2-4}\right)\\
&=\lim_{x\to 2} \left(\frac{6-x-x+2}{x^2-4}\right)\\
&=\lim_{x\to 2} \left(\frac{4-2x}{(x-2)(x+2)}\right)\\
&=\lim_{x\to 2} \left(\frac{2(2-x)}{(x-2)(x+2)}\right)\\
&=\lim_{x\to 2} \left(\frac{-2}{(x+2)}\right)\\
&=\frac{-2}{2+2}\\
&=\frac{-2}{4}\\
&=-\frac{1}{2}\\
\end{align}\)

\(\begin{align}
\lim_{x\to 1} \left(\frac{1}{1-x}-\frac{2}{x-x^3}\right)&=\lim_{x\to 1} \left(\frac{1}{1-x}-\frac{2}{x(1-x^2)}\right)\\
&=\lim_{x\to 1} \left(\frac{1}{1-x}-\frac{2}{x(1-x)(1+x))}\right)\\
&=\lim_{x\to 1} \left(\frac{(x+x^2)-2}{x(1-x)(1+x)}\right)\\
&=\lim_{x\to 1} \left(\frac{x^2+x-2}{x(1-x)(1+x)}\right)\\
&=\lim_{x\to 1} \left(\frac{(x+2)(x-1)}{x(1-x)(1+x)}\right)\\
&=\lim_{x\to 1} \left(\frac{-(x+2)}{x(1+x)}\right)\\
&=\frac{-(1+2)}{1(1+1)}\\
&=-\frac{3}{2}\\
\end{align}\)

\(\begin{align}
\lim_{x\to 1} \frac{x-1}{\sqrt{x+3}-2}&=\lim_{x\to 1} \frac{x-1}{\sqrt{x+3}-2}\times\frac{\sqrt{x+3}+2}{\sqrt{x+3}+2}\\
&=\lim_{x\to 1} \frac{(x-1)(\sqrt{x+3}+2)}{x+3-4}\\
&=\lim_{x\to 1} \frac{(x-1)(\sqrt{x+3}+2)}{x-1}\\
&=\lim_{x\to 1} (\sqrt{x+3}+2)\\
&=(\sqrt{1+3}+2)\\
&=(\sqrt{4}+2)\\
&=2+2\\
&=4\\
\end{align}\)

\(\begin{align}
\lim_{x\to 3} \frac{x-\sqrt{2x+3}}{9-x^2}&=\lim_{x\to 3} \frac{x-\sqrt{2x+3}}{9-x^2}\times \frac{x+\sqrt{2x+3}}{x+\sqrt{2x+3}}\\
&=\lim_{x\to 3} \frac{x^2-{(2x+3)}}{(9-x^2)(x+\sqrt{2x+3})}\\
&=\lim_{x\to 3} \frac{x^2-2x-3}{(9-x^2)(x+\sqrt{2x+3})}\\
&=\lim_{x\to 3} \frac{(x-3)(x+1)}{(3-x)(3+x)(x+\sqrt{2x+3})}\\
&=\lim_{x\to 3} \frac{-(x+1)}{(3+x)(x+\sqrt{2x+3})}\\
&=\frac{-(3+1)}{(3+3)(3+\sqrt{6+3})}\\
&=\frac{-4}{(6)(3+\sqrt{9})}\\
&=\frac{-4}{(6)(3+3)}\\
&=\frac{-4}{(6)(6)}\\
&=\frac{-4}{36}\\
&=-\frac{1}{9}\\
\end{align}\)

\(\begin{align}
\lim_{x\to 0} \frac{x^2}{1-\sqrt{1+x^2}}&=\lim_{x\to 0} \frac{x^2}{1-\sqrt{1+x^2}}\times \frac{1+\sqrt{1+x^2}}{1+\sqrt{1+x^2}}\\
&=\lim_{x\to 0} \frac{x^2(1+\sqrt{1+x^2})}{(1-\sqrt{1+x^2})(1+\sqrt{1+x^2})}\\
&=\lim_{x\to 0} \frac{x^2(1+\sqrt{1+x^2})}{1-({1+x^2})}\\
&=\lim_{x\to 0} \frac{x^2(1+\sqrt{1+x^2})}{-x^2}\\
&=\lim_{x\to 0}-(1+\sqrt{1+x^2})\\
&=-(1+\sqrt{1+0^2})\\
&=-(1+1)\\
&=-2\\
\end{align}\)

\(\begin{align}
\lim_{x\to 2} \left(\frac{2}{x^2-4}-\frac{3}{x^2+2x-8}\right)&=\lim_{x\to 1} \left(\frac{2}{(x-2)(x+2)}-\frac{3}{(x+4)(x-2)}\right)\\
&=\lim_{x\to 2} \left(\frac{2(x+4)-3(x+2)}{(x-2)(x+2)(x+4)}\right)\\
&=\lim_{x\to 2} \left(\frac{(2x+8-3x-6)}{(x-2)(x+2)(x+4)}\right)\\
&=\lim_{x\to 2} \left(\frac{(-x+2)}{(x-2)(x+2)(x+4)}\right)\\
&=\lim_{x\to 2} \left(\frac{-(x-2)}{(x-2)(x+2)(x+4)}\right)\\
&=\lim_{x\to 2} \left(\frac{-1}{(x+2)(x+4)}\right)\\
&=\frac{-1}{(2+2)(2+4)}\\
&=-\frac{1}{(4)(6)}\\
&=-\frac{1}{24}\\
\end{align}\)