Obliczanie granic - przykłady

\[\lim_{n \to \infty} \frac{1}{n}+3=0+3=3\]

\[\lim_{n \to \infty} \frac{1}{n}-\sqrt{2}=0-\sqrt{2}=-\sqrt{2}\]

\[\begin{split} \lim_{n \to \infty} \frac{1}{5n}+21 &=\lim_{n \to \infty} \frac{1}{5}\cdot \frac{1}{n}+21=\\[16pt] &=\frac{1}{5}\cdot 0+21=\\[16pt] &=0+21=21 \end{split}\]

\[\lim_{n \to \infty} \frac{1}{n^2}+\frac{2}{n^3}-\frac{100}{n^6}=0+0-0=0\]

\[\begin{split}&\lim_{n \to \infty} n\left(\sqrt{2n^2+1}-\sqrt{2n^2-1}\right)=\\[16pt] &=\lim_{n \to \infty} \frac{n\left(2n^2+1-(2n^2-1)\right)}{\left(\sqrt{2n^2+1}+\sqrt{2n^2-1}\right)}=\\[16pt] &=\lim_{n \to \infty} \frac{2n}{\left(\sqrt{2n^2+1}+\sqrt{2n^2-1}\right)}\frac{:n}{:n}=\\[16pt] &=\lim_{n \to \infty} \dfrac{\dfrac{2n}{n}}{\sqrt{\dfrac{2n^2}{n^2}+\dfrac{1}{n^2}}+\sqrt{\dfrac{2n^2}{n^2}-\dfrac{1}{n^2}}}=\\[16pt] &=\lim_{n \to \infty} \dfrac{2}{\sqrt{2+\dfrac{1}{n^2}}+\sqrt{2-\dfrac{1}{n^2}}}=\\[16pt] &=\frac{2}{2\sqrt{2}}=\frac{\sqrt{2}}{2}\end{split}\]

\[\begin{split} &\lim_{n \to \infty} n\left(\sqrt{7n^2+3}-\sqrt{7n^2-3}\right)=\\[16pt] &=\lim_{n \to \infty} \frac{n\left(7n^2+3-(7n^2-3)\right)}{\left(\sqrt{7n^2+3}+\sqrt{7n^2-3}\right)}=\\[16pt] &=\lim_{n \to \infty} \frac{6n}{\left(\sqrt{7n^2+3}+\sqrt{7n^2-3}\right)}\frac{:n}{:n}=\\[16pt] &=\lim_{n \to \infty} \dfrac{\dfrac{6n}{n}}{\sqrt{\dfrac{7n^2}{n^2}+\dfrac{3}{n^2}}+\sqrt{\dfrac{7n^2}{n^2}-\dfrac{3}{n^2}}}=\\[16pt] &=\lim_{n \to \infty} \dfrac{6}{\sqrt{7+\dfrac{3}{n^2}}+\sqrt{7-\dfrac{3}{n^2}}}=\\[16pt] &=\frac{6}{2\sqrt{7}}=\frac{3}{\sqrt{7}}=\frac{3\sqrt{7}}{7} \end{split}\]

\[\begin{split} &\lim_{n \to \infty} \frac{5n^6-3n^4+2}{5-9n^6}=\\[15pt] &=\lim_{n \to \infty} \frac{5n^6-3n^4+2}{5-9n^6}\ \frac{:n^6}{:n^6}=\\[15pt] &=\lim_{n \to \infty} \dfrac{\dfrac{5n^6}{n^6}-\dfrac{3n^4}{n^6}+\dfrac{2}{n^6}}{\dfrac{5}{n^6}-\dfrac{9n^6}{n^6}}=\\[15pt] &=\lim_{n \to \infty} \dfrac{5-\dfrac{3}{n^2}+\dfrac{2}{n^6}}{\dfrac{5}{n^6}-9}=\\[15pt] &=-\frac{5}{9} \end{split}\]

\[ \begin{split} &\lim_{n \to \infty} \sqrt{n^2+4n+1}-\sqrt{n^2+2n}=\\[12pt] &=\lim_{n \to \infty} \frac{n^2+4n+1-n^2-2n}{\sqrt{n^2+4n+1}+\sqrt{n^2+2n}}=\\[16pt] &=\lim_{n \to \infty} \frac{2n+1}{\sqrt{n^2+4n+1}+\sqrt{n^2+2n}}\frac{:n}{:n}=\\[16pt] &=\lim_{n \to \infty} \dfrac{\dfrac{2n}{n}+\dfrac{1}{n}}{\sqrt{\dfrac{n^2}{n^2}+\dfrac{4n}{n^2}+\dfrac{1}{n^2}}+\sqrt{\dfrac{n^2}{n^2}+\dfrac{2n}{n^2}}}=\\[16pt] &=\lim_{n \to \infty} \dfrac{2+\dfrac{1}{n}}{\sqrt{1+\dfrac{4}{n}+\dfrac{1}{n^2}}+\sqrt{1+\dfrac{2}{n}}}=\\[16pt] &=\frac{2}{\sqrt{1}+\sqrt{1}}=\frac{2}{2}=1 \end{split} \]

\[ \begin{split} &\lim_{n \to \infty} \frac{1-2+3-4+...-2n}{\sqrt{n^2+1}}=\\[16pt] &=\lim_{n \to \infty} \frac{\left(1+3+...+(2n-1)\right)-(2+4+...+2n)}{\sqrt{n^2+1}}=\\[16pt] &\{\text{w liczniku mamy dwie sumy ciągów arytmetycznych}\}\\[16pt] &=\lim_{n \to \infty} \frac{\dfrac{(2n-1)+1}{2}\cdot n-\dfrac{2n+2}{2}\cdot n}{\sqrt{n^2+1}}=\\[16pt] &=\lim_{n \to \infty} \frac{n^2-n^2-n}{\sqrt{n^2+1}}=\\[16pt] &=\lim_{n \to \infty} \frac{-n}{\sqrt{n^2+1}}\frac{:n}{:n}=\\[16pt] &=\lim_{n \to \infty} \dfrac{-\dfrac{n}{n}}{\sqrt{\dfrac{n^2}{n^2}+\dfrac{1}{n^2}}}=\\[16pt] &=\lim_{n \to \infty} \dfrac{-1}{\sqrt{1+\dfrac{1}{n^2}}}=\\[16pt] &=\frac{-1}{\sqrt{1}}=-1 \end{split} \]

W liczniku mamy sumę ciągu geometrycznego: \[ 1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2^n}=\frac{1}{1-\frac{1}{2}}=2 \] W mianowniku również mamy sumę ciągu geometrycznego: \[ 1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^n}=\frac{1}{1-\frac{1}{3}}=\frac{3}{2} \] Zatem mamy: \[ \lim_{n \to \infty} \dfrac{1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2^n}}{1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^n}}=\dfrac{2}{\frac{3}{2}}=\frac{4}{3} \]

\[\begin{split} &\lim_{n \to \infty} \frac{\sqrt{n}}{\sqrt{n+\sqrt{n+\sqrt{n}}}}=\\[16pt] &=\lim_{n \to \infty} \frac{\sqrt{n}}{\sqrt{n+\sqrt{n+\sqrt{n}}}}\ \frac{:\sqrt{n}}{:\sqrt{n}}=\\[16pt] &=\lim_{n \to \infty} \dfrac{\sqrt{\dfrac{n}{n}}}{\sqrt{\dfrac{n}{n}+\sqrt{\dfrac{n}{n^2}+\sqrt{\dfrac{n}{n^4}}}}}=\\[16pt] &=\lim_{n \to \infty} \dfrac{\sqrt{1}}{\sqrt{1+\sqrt{\dfrac{1}{n}+\sqrt{\dfrac{1}{n^3}}}}}=\\[16pt] &=\frac{1}{\sqrt{1}}=1 \end{split}\]

\[ \begin{split} &\lim_{n \to \infty} \sqrt{2}\cdot \sqrt[4]{2}\cdot \sqrt[8]{2}\cdot ...\cdot \sqrt[2^n]{2}=\\[16pt] &=\lim_{n \to \infty} 2^\tfrac{1}{2}\cdot 2^\tfrac{1}{4}\cdot ...\cdot 2^\tfrac{1}{2^n}=\\[16pt] &=\lim_{n \to \infty} 2^{\tfrac{1}{2}+\tfrac{1}{4}+\tfrac{1}{8}+...+\tfrac{1}{2^n}}=\\[16pt] &= 2^{\tfrac{\tfrac{1}{2}}{1-\tfrac{1}{2}}}=\\[16pt] &= 2^{\tfrac{1}{2}\cdot \tfrac{2}{1}}=\\[16pt] &= 2^1=2 \end{split} \]

\[ \begin{split} &\lim_{n \to \infty} \left(\sqrt{n+6\sqrt{n}+1}-\sqrt{n}\right)=\\[16pt] &=\lim_{n \to \infty} \left(\sqrt{n+6\sqrt{n}+1}-\sqrt{n}\right)\cdot \frac{\sqrt{n+6\sqrt{n}+1}+\sqrt{n}}{\sqrt{n+6\sqrt{n}+1}+\sqrt{n}}=\\[16pt] &=\lim_{n \to \infty} \frac{n+6\sqrt{n}+1-n}{\sqrt{n+6\sqrt{n}+1}+\sqrt{n}}=\\[16pt] &=\lim_{n \to \infty} \frac{6\sqrt{n}+1}{\sqrt{n+6\sqrt{n}+1}+\sqrt{n}}\frac{:\sqrt{n}}{:\sqrt{n}}=\\[16pt] &=\lim_{n \to \infty} \dfrac{6\sqrt{\dfrac{n}{n}}+\sqrt{\dfrac{1}{n}}}{\sqrt{\dfrac{n}{n}+6\sqrt{\dfrac{n}{n^2}}+\dfrac{1}{n}}+\sqrt{\dfrac{n}{n}}}=\\[16pt] &=\lim_{n \to \infty} \dfrac{6+\sqrt{\dfrac{1}{n}}}{\sqrt{1+6\sqrt{\dfrac{1}{n}}+\dfrac{1}{n}}+1}=\\[16pt] &= \frac{6+\sqrt{0}}{\sqrt{1+0+0}+1}=\frac{6}{2}=3 \end{split} \]

W liczniku pod pierwiastkiem mamy sumę ciągu arytmetycznego, zatem: \[ \begin{split} &\lim_{n \to \infty} \frac{\sqrt{1+2+3+...+n}}{n}=\\[16pt] &=\lim_{n \to \infty} \dfrac{\sqrt{\dfrac{1+n}{2}\cdot n}}{n}=\\[16pt] &=\lim_{n \to \infty} \dfrac{\sqrt{\dfrac{n+n^2}{2}}}{n}\cdot \dfrac{\dfrac{1}{n}}{\dfrac{1}{n}}=\\[16pt] &=\lim_{n \to \infty} \dfrac{\sqrt{\dfrac{n+n^2}{2n^2}}}{1}=\\[16pt] &=\lim_{n \to \infty} \sqrt{\frac{1}{2n}+\frac{1}{2}}=\\[16pt] &=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2} \end{split} \]

\[ \begin{split} &\lim_{n \to \infty} \frac{{\sqrt{n^2+\sqrt{n+1}}}-\sqrt{n^2-\sqrt{n-1}}}{\sqrt{n+1}-\sqrt{n}}=\\[16pt] &=\lim_{n \to \infty} \frac{{\sqrt{n^2+\sqrt{n+1}}}-\sqrt{n^2-\sqrt{n-1}}}{\sqrt{n+1}-\sqrt{n}}\cdot \frac{({\sqrt{n^2+\sqrt{n+1}}}+\sqrt{n^2-\sqrt{n-1}})(\sqrt{n+1}+\sqrt{n})}{({\sqrt{n^2+\sqrt{n+1}}}+\sqrt{n^2-\sqrt{n-1}})(\sqrt{n+1}+\sqrt{n})} =\\[16pt] &=\lim_{n \to \infty} \frac{(n^2+\sqrt{n+1}-n^2+\sqrt{n-1})(\sqrt{n+1}+\sqrt{n})}{\left(\sqrt{n^2+\sqrt{n+1}}+\sqrt{n^2-\sqrt{n-1}}\right)(n+1-n)}=\\[16pt] &=\lim_{n \to \infty} \frac{(\sqrt{n+1}+\sqrt{n-1})(\sqrt{n+1}+\sqrt{n})}{\left(\sqrt{n^2+\sqrt{n+1}}+\sqrt{n^2-\sqrt{n-1}}\right)}=\\[16pt] &=\lim_{n \to \infty} \frac{(n+1+\sqrt{n^2+n}+\sqrt{n^2-1}+\sqrt{n^2-n})}{\left(\sqrt{n^2+\sqrt{n+1}}+\sqrt{n^2-\sqrt{n-1}}\right)}\frac{:n}{:n}=\\[16pt] &=\lim_{n \to \infty} \frac{1+\frac{1}{n}+\sqrt{1+\frac{1}{n}}+\sqrt{1-\frac{1}{n^2}}+\sqrt{1-\frac{1}{n}}}{\sqrt{1+\sqrt{\frac{1}{n^3}+\frac{1}{n^4}}}+\sqrt{1-\sqrt{\frac{1}{n^3}-\frac{1}{n^4}}}}=\\[16pt] &=\frac{1+0+1+1+1}{\sqrt{1}+\sqrt{1}}=\frac{4}{2}=2 \end{split} \]

\[ \begin{split} &\lim_{n \to \infty} \frac{(n+2)!+(n+1)!}{(n+2)!-(n+1)!}=\\[16pt] &=\lim_{n \to \infty} \frac{(n+1)!\cdot (n+2)+(n+1)!}{(n+1)!\cdot (n+2)-(n+1)!}=\\[16pt] &=\lim_{n \to \infty} \frac{(n+1)!\cdot (n+2+1)}{(n+1)!\cdot (n+2-1)}=\\[16pt] &=\lim_{n \to \infty} \frac{(n+1)!\cdot (n+3)}{(n+1)!\cdot (n+1)}=\\[16pt] &=\lim_{n \to \infty} \frac{n+3}{n+1}\cdot \frac{\frac{1}{n}}{\frac{1}{n}}=\\[16pt] &=\lim_{n \to \infty} \frac{1+\frac{3}{n}}{1+\frac{1}{n}}=\\[16pt] &=\frac{1}{1}=1 \end{split} \]

\[ \begin{split} \lim_{n \to \infty} \frac{7^n+5^n}{5^n+3^n}&=\lim_{n \to \infty} \frac{7^n\left(1+\left(\dfrac{5}{7}\right)^n\right)}{5^n\left(1+\left(\dfrac{3}{5}\right)^n\right)}=\\[16pt] &=\lim_{n \to \infty} \left(\frac{7}{5}\right)^n=\infty \end{split} \]

\[ \begin{split} &\lim_{n \to \infty} n\left(\ln (n+1)-\ln n\right)=\\[16pt] &=\lim_{n \to \infty} n\left(\ln \frac{n+1}{n}\right)=\\[16pt] &=\lim_{n \to \infty} \ln \left(\frac{n+1}{n}\right)^n=\\[16pt] &=\lim_{n \to \infty} \ln \left(1+\frac{1}{n}\right)^n=\\[16pt] &=\lim_{n \to \infty} \ln e=1 \end{split} \]

\[ \begin{split} &\lim_{n \to \infty} \frac{\log_2(n+1)}{\log_3(n+1)}=\\[16pt] &=\lim_{n \to \infty} \frac{\log_2(n+1)}{\frac{\log_2(n+1)}{\log_23}}=\\[16pt] &=\lim_{n \to \infty} \log_23=\\[16pt] &=\log_23 \end{split} \]

\[ \begin{split} &\lim_{n \to \infty} (1+2^n-3^n)=\\[16pt] &=\lim_{n \to \infty} 3^n\left(\frac{1}{3^n}+\left(\frac{2}{3}\right)^n-1\right)=\\[16pt] &=-\lim_{n \to \infty} 3^n=-\infty \end{split} \]

\[ \begin{split} &\lim_{n \to \infty} \left(\frac{n+5}{n}\right)^n=\\[16pt] &=\lim_{n \to \infty} \left(1+\frac{5}{n}\right)^n=\\[16pt] &=\lim_{n \to \infty}\left(1+\frac{1}{\frac{n}{5}}\right)^n=\\[16pt] &=\lim_{n \to \infty} \left[\left(1+\frac{1}{\frac{n}{5}}\right)^\dfrac{n}{5}\right]^5=\\[16pt] &=e^5 \end{split} \]

\[ \begin{split} &\lim_{n \to \infty} \left(1-\frac{1}{n^2}\right)^n=\\[16pt] &=\lim_{n \to \infty} \left(\left(1-\frac{1}{n^2}\right)^{n^2}\right)^{\frac{1}{n}}=\\[16pt] &=e^0=1 \end{split} \]

\[ \begin{split} &\lim_{n \to \infty} \left(\frac{n^2+6}{n^2}\right)^{n^2}=\\[6pt] &=\lim_{n \to \infty} \left(1+\frac{6}{n^2}\right)^{n^2}=\\[6pt] &=\lim_{n \to \infty} \left[\left(1+\frac{6}{n^2}\right)^{\dfrac{n^2}{6}}\right]^6=\\[6pt] &=e^6 \end{split} \]

\[ \lim_{x \to {-3}} (x^2+3x+7)=(-3)^2+3\cdot (-3)+7=7 \]

\[ \lim_{x \to 4}\frac{\sqrt{x^2-16}}{4x+2} =\frac{\sqrt{4^2-16}}{4\cdot 4+2}=\frac{0}{18}=0 \]

\[ \begin{split} &\lim_{x \to 3}\frac{x^2-6x+9}{x^2-9} =\\[15pt] &=\lim_{x \to 3}\frac{(x-3)^2}{(x-3)(x+3)}=\\[15pt] &=\lim_{x \to 3}\frac{x-3}{x+3}=\\[15pt] &=\frac{0}{6}=0 \end{split} \]

\[ \begin{split} &\lim_{x \to 1}\frac{1-x^2}{\left(1-\sqrt{x}\right)} =\\[16pt] &=\lim_{x \to 1}\frac{(1-x)(1+x)}{\left(1-\sqrt{x}\right)} =\\[16pt] &=\lim_{x \to 1}\frac{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)(1+x)}{\left(1-\sqrt{x}\right)} =\\[16pt] &=\lim_{x \to 1} \left(1+\sqrt{x}\right)(1+x)=\\[16pt] &=2\cdot 2=4 \end{split} \]

\[ \begin{split} &\lim_{z \to -2}\frac{z(z^2+4z+4)}{(z+2)(z-3)}=\\[16pt] &=\lim_{z \to -2}\frac{z(z+2)^2}{(z+2)(z-3)}=\\[16pt] &=\lim_{z \to -2}\frac{z(z+2)}{z-3} =\\[16pt] &=\frac{0}{-5}=0 \end{split} \]

\[\begin{split} &\lim_{n \to \infty} \frac{2x^2-1}{7x^2+2x}=\\[16pt] &=\lim_{n \to \infty} \frac{2x^2-1}{7x^2+2x}\cdot \frac{\frac{1}{x^2}}{\frac{1}{x^2}}=\\[16pt] &=\lim_{n \to \infty} \frac{2-\dfrac{1}{x^2}}{7+\dfrac{2}{x}}=\\[16pt] &=\frac{2}{7} \end{split}\]

\[\begin{split} &\lim_{n \to -\infty}\frac{1+\sqrt{2x^2-1}}{x}=\\[16pt] &=\lim_{n \to -\infty}\frac{1+\sqrt{2x^2-1}}{x}\cdot \frac{\frac{1}{|x|}}{\frac{1}{|x|}}=\\[16pt] &=\lim_{n \to -\infty}\dfrac{\dfrac{1}{|x|}+\sqrt{\dfrac{2x^2}{|x^2|}-\dfrac{1}{|x^2|}}}{\dfrac{x}{|x|}}=\\[16pt] &=\lim_{n \to -\infty}\dfrac{\dfrac{1}{|x|}+\sqrt{\dfrac{2x^2}{x^2}-\dfrac{1}{x^2}}}{-1}=\\[16pt] &=\lim_{n \to -\infty}\dfrac{\dfrac{1}{|x|}+\sqrt{2-\dfrac{1}{x^2}}}{-1}=\\[16pt] &=\dfrac{0+\sqrt{2-0}}{-1}=\\[16pt] &=-\sqrt{2} \end{split}\]

\[ \begin{split} &\lim_{x \to 2}\left(\frac{1}{x-2}-\frac{4}{x^2-4}\right)=\\[16pt] &=\lim_{x \to 2}\left(\frac{x+2}{(x-2)(x+2)}-\frac{4}{(x-2)(x+2)}\right)=\\[16pt] &=\lim_{x \to 2}\frac{(x-2)}{(x-2)(x+2)}=\\[16pt] &=\lim_{x \to 2}\frac{1}{x+2}=\frac{1}{4} \end{split} \]

\[ \begin{split} &\lim_{x \to 2} \frac{x^4-8x^2+16}{(x-2)(x-3)}=\\[16pt] &=\lim_{x \to 2}\frac{(x^2-4)^2}{(x-2)(x-3)}=\\[16pt] &=\lim_{x \to 2}\frac{(x-2)^2(x+2)^2}{(x-2)(x-3)}=\\[16pt] &=\lim_{x \to 2}\frac{(x-2)(x+2)^2}{x-3}=\\[16pt] &=\frac{0}{-1}=0 \end{split} \]

\[ \begin{split} &\lim_{x \to \infty} \frac{\sqrt{1+x}+3x}{\sqrt{1+x^2}}=\\[16pt] &=\lim_{x \to \infty} \frac{\sqrt{1+x}+3x}{\sqrt{1+x^2}}\cdot \frac{\frac{1}{x}}{\frac{1}{x}}=\\[16pt] &=\lim_{x \to \infty} \frac{\sqrt{\dfrac{1}{x^2}+\dfrac{x}{x^2}}+\dfrac{3x}{x}}{\sqrt{\dfrac{1}{x^2}+\dfrac{x^2}{x^2}}}=\\[16pt] &=\lim_{x \to \infty} \frac{\sqrt{\dfrac{1}{x^2}+\dfrac{1}{x}}+3}{\sqrt{\dfrac{1}{x^2}+1}}=\\[16pt] &=\frac{3}{\sqrt{1}}=3 \end{split} \]

\[ \begin{split} &\lim_{x \to -1} \frac{x^4+3x^2-4}{x+1}=\\[16pt] &=\lim_{x \to -1}\frac{(x^2+4)(x^2-1)}{x+1}=\\[16pt] &=\lim_{x \to -1}\frac{(x^2+4)(x-1)(x+1)}{x+1}=\\[16pt] &=\lim_{x \to -1}(x^2+4)(x-1)=\\[16pt] &=5\cdot (-2)=-10 \end{split} \]

\[ \lim_{x \to 0} \frac{\sin 2x}{x}=\lim_{x \to 0}\frac{2\sin 2x}{2x}=2\lim_{x \to 0}\frac{\sin 2x}{2x}=2 \]

\[ \begin{split} &\lim_{x \to -1}\frac{\sin (x+1)}{1-x^2}=\\[16pt] &=\lim_{x \to -1}\frac{\sin (x+1)}{(1-x)(1+x)}=\\[16pt] &=\lim_{x \to -1}\frac{\sin (x+1)}{1+x}\cdot \lim_{x \to -1}\frac{1}{1-x}=\\[16pt] &=1\cdot \frac{1}{2}=\frac{1}{2} \end{split} \]

\[ \begin{split} &\lim_{x \to \dfrac{\pi }{4}}\frac{\sqrt{\sin x}-\sqrt{\cos x}}{\sin x-\cos x} =\\[16pt] &=\lim_{x \to \dfrac{\pi }{4}}\frac{(\sqrt{\sin x}-\sqrt{\cos x})}{(\sqrt{\sin x}-\sqrt{\cos x})(\sqrt{\sin x}+\sqrt{\cos x})}=\\[16pt] &=\lim_{x \to \dfrac{\pi }{4}}\frac{1}{\sqrt{\sin x}+\sqrt{\cos x}}=\\[16pt] &=\frac{1}{\sqrt{\frac{\sqrt{2}}{2}}+\sqrt{\frac{\sqrt{2}}{2}}}=\frac{1}{\sqrt{2\sqrt{2}}} \end{split} \]