جد الغاية لكل مما يأتي: $\underset{\mathrm{x}\to 0}{lim}[\frac{3\mathrm{x}}{\sin 2\mathrm{x}}+\frac{1-\cos 6\mathrm{x}}{{\sin }^2\mathrm{x}}]$

$\begin{array}{rl}1-\cos 6x=& 1-(1-2{\sin }^{2}3x)\\ & =1-1+2{\sin }^{2}3x\\ & =2{\sin }^{2}3x\\ & =\underset{x\to 0}{lim}\frac{3x}{\sin 2x}+\underset{x\to 0}{lim}\frac{2{\sin }^{2}3x}{{\sin }^{2}x}\\ & =\frac{3}{2}\underset{x\to 0}{lim}\frac{2x}{\sin 2x}+2\underset{x\to 0}{lim}\frac{{\sin }^{2}3x}{{\sin }^{2}x}\\ & =\frac{3}{2}\underset{x\to 0}{lim}\frac{1}{\sin 2x}+2\underset{x\to 0}{lim}\frac{\frac{{\sin }^{2}3x}{x^2}}{\frac{{\sin }^{2}x}{x^2}}\end{array}$

$\begin{array}{rl}& =\frac{3}{2}\times \frac{1}{1}+9\cdot 2\underset{x\to 0}{lim}\frac{\frac{{\sin }^{2}3x}{9x^2}}{\frac{{\sin }^{2}x}{x^2}}\\ & =\frac{3}{2}+18\underset{x\to 0}{lim}\frac{{\left[\frac{\sin 3x}{3x}\right]}^2}{{\left[\frac{\sin x}{x}\right]}^2}=\frac{3}{2}+18\cdot \frac{1}{1}=\frac{3}{2}+18\\ & =\frac{3+36}{2}=\frac{39}{2}\end{array}$

تمارين (1-5)

(1)- جد الغاية لكل مما يأتي:

أ- $\underset{x\to 3}{lim}\frac{(x^2-x-6)}{(x-3)}$

$\underset{\mathrm{x}\to 3}{lim}\frac{(\mathrm{x}-3)(\mathrm{x}+2)}{(\mathrm{x}-3)}=\underset{\mathrm{x}\to 3}{lim}\mathrm{x}+2=3^{}+2=5$

ب- $\underset{\mathrm{x}\to 1}{lim}(3\mathrm{x}-4)$

$\underset{\mathrm{x}\to 1}{lim}(3\mathrm{x}-4)=3\times 1-4=-1$

ج- $\underset{\mathrm{x}\to 1}{lim}\frac{({\mathrm{x}}^3-1)}{(2\mathrm{x}-2)}$

$\begin{array}{rl}\underset{x\to 1}{lim}\frac{(x^3-1)}{(2x-2)}=& \underset{x\to 1}{lim}\frac{(x-1)(x^2+x+1)}{2(x-1)}=\underset{x\to 1}{lim}\frac{(x^2+x+1)}{2}\\ & =\frac{1^2+1+1}{2}=\frac{3}{2}=1.5\end{array}$

د- $\underset{\mathrm{x}\to 4}{lim}\frac{{\mathrm{x}}^2-16}{\sqrt{\mathrm{x}+5}-3}, \{\mathrm{x}:\mathrm{x}\ge -5\}/\left\{4\right\}$

$\begin{array}{rl}\underset{x\to 4}{lim}\frac{x^2-16}{\sqrt{x+5}-3}\times \frac{\sqrt{x+5}+3}{\sqrt{x+5}+3}& =\underset{x\to 4}{lim}\frac{(x^2-16)(\sqrt{x+5}+3)}{(x+5-9)}\\ =\underset{x\to 4}{lim}\frac{(x-4)(x+4)(\sqrt{x+5}+3)}{(x-4)}& =(4+4)(\sqrt{x+5}+3)\\ & =\left(8\right)(3+3)=\left(8\right)\times \left(6\right)=48\end{array}$

هـ- $\underset{\mathrm{x}\to 3}{lim}\frac{{\mathrm{x}}^2-2\mathrm{x}-3}{{\mathrm{x}}^2-9}$

$\underset{\mathrm{x}\to 4}{lim}\frac{(\mathrm{x}-3)(\mathrm{x}+1)}{(\mathrm{x}-3)(\mathrm{x}+3)}=\underset{\mathrm{x}\to 4}{lim}\frac{(\mathrm{x}+1)}{(\mathrm{x}+3)}=\frac{4+1}{4+3}=\frac{5}{7}$

(2)- إذا كان: $\mathrm{f}:\mathrm{R}\to \mathrm{R}$

جد $\underset{\mathrm{x}\to 1}{lim}\mathrm{f}\left(\mathrm{x}\right)$ حيث $\mathrm{f}\left(\mathrm{x}\right)=|\mathrm{x}-1|$

$\mathrm{f}\left(\mathrm{x}\right)=\{\begin{array}{lll}{\mathrm{x}}^2-1& & \mathrm{x}\ge 1\\ -(\mathrm{x}-1)& & \mathrm{x}<1\end{array}$

$\begin{array}{c}\underset{\mathrm{x}\to 1}{lim}\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to 1}{lim}\mathrm{x}-1=1-1=0={\mathrm{L}}_1\\ \underset{\mathrm{x}\to 1}{lim}\mathrm{F}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to 1}{lim}-(1-1)=0={\mathrm{L}}_2\end{array}$

الغاية موجودة لأن $\mathrm{L}1=\mathrm{L}2$

(3)- $\mathrm{f}:\mathrm{R}\to \mathrm{R}$

$\mathrm{f}\left(\mathrm{x}\right)=\{\begin{array}{ll}5-{\mathrm{x}}^2& \mathrm{x}>-1\\ {\mathrm{x}}^2+3& \mathrm{x}<-1\\ 4& \mathrm{x}=-1\end{array}$

أ- ارسم المخطط البياني لهذه الدالة.

ب- هل للدالة غاية عند 1- بين ذلك؟

$\begin{array}{rl}& \underset{\mathrm{x}\to 1}{lim}5-{\mathrm{x}}^2=5-(-1{)}^2=5-1=4={\mathrm{L}}_1\\ & \underset{\mathrm{x}\to -1}{lim}{\mathrm{x}}^2+3=(-1{)}^2+3=1+3=4={\mathrm{L}}_2\end{array}$

الغاية موجودة لأن $\mathrm{L}1=\mathrm{L}2$

ج- جد $\underset{\mathrm{x}\to \sqrt{2}}{lim}\mathrm{f}\left(\mathrm{x}\right)$

$\underset{\mathrm{x}\to \sqrt{2}}{lim}5-{\mathrm{x}}^2=5-(\sqrt{2}{)}^2=5-2=3$

(4)- $\mathrm{f}\left(\mathrm{x}\right)=\{\begin{array}{ll}{\mathrm{x}}^2+\mathrm{a}& \mathrm{x}>-1\\ 6& \mathrm{x}=-1\\ 4\mathrm{x}+\mathrm{b}& \mathrm{x}<-1\end{array}$

إذا كانت $\underset{\mathrm{x}\to -1}{lim}\mathrm{f}\left(\mathrm{x}\right)=3$ جد قيمة $\mathrm{a},\mathrm{b}\in \mathrm{R}$

$$\begin{gathered} \underset{\mathrm{x}\to -1}{lim}\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to -1}{lim}\mathrm{f}\left(\mathrm{x}\right)=3 \\ \underset{\mathrm{x}\to -1}{lim}\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to -1}{lim}{\mathrm{x}}^2+\mathrm{a}\to (-1{)}^2+\mathrm{a}=3\to 1+\mathrm{a}=3\to \mathrm{a}=2 \\ \underset{\mathrm{x}\to -1}{lim}\mathrm{f}(\mathrm{x})=\underset{\mathrm{x}\to -1}{lim}4\mathrm{x}+\mathrm{b}\to 4(-1)+\mathrm{b}=3\to -4+\mathrm{b}=3\to \mathrm{b}=7 \end{gathered}$$

(5)- إذا كان:

$\begin{array}{rl}& \mathrm{g}\left(\mathrm{x}\right)=3{\mathrm{x}}^2+2\mathrm{x}-3\\ & \mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2+6\end{array}$

جد:

$\begin{array}{rl}& \underset{\mathrm{x}\to 0}{lim}(\mathrm{g}/\mathrm{f})\left(\mathrm{x}\right)\\ & \underset{\mathrm{x}\to 0}{lim}(\mathrm{g}.\mathrm{f})\left(\mathrm{x}\right)\end{array}$

$\underset{\mathrm{x}\to 0}{lim}\frac{\mathrm{g}\left(\mathrm{x}\right)}{\mathrm{f}\left(\mathrm{x}\right)}=\frac{\underset{\mathrm{x}\to 0}{lim}(3{\mathrm{x}}^2+2\mathrm{x}-3)}{\underset{\mathrm{x}\to 0}{lim}({\mathrm{x}}^2+6)}=\frac{3(0{)}^2+2(0)-3}{(0{)}^2+6}=\frac{-3}{6}=-\frac{1}{2}$

$\begin{array}{rl}\underset{\mathrm{x}\to 0}{lim}\mathrm{g}\left(\mathrm{x}\right)\cdot \mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to 0}{lim}\mathrm{g}\left(\mathrm{x}\right)\cdot \underset{\mathrm{x}\to 0}{lim}\mathrm{f}\left(\mathrm{x}\right)& =\underset{\mathrm{x}\to 0}{lim}(3{\mathrm{x}}^2+2\mathrm{x}-3)\cdot \underset{\mathrm{x}\to 0}{lim}({\mathrm{x}}^2+6)\\ & =-3\times 6=-18\end{array}$

(6)- جد الغاية لكل مما يأتي:

أ- $\underset{\mathrm{x}\to 0}{lim}\frac{\tan {\mathrm{x}}^2}{{\sin }^2\mathrm{x}}$

$\begin{array}{rl}\underset{\mathrm{x}\to 0}{lim}\frac{{\tan }^2\mathrm{x}}{{\sin }^2\mathrm{x}}& =\underset{\mathrm{x}\to 0}{lim}\frac{\frac{{\sin }^2\mathrm{x}}{{\cos }^2\mathrm{x}}}{{\sin }^2\mathrm{x}}=\underset{\mathrm{x}\to 0}{lim}\frac{{\sin }^2\mathrm{x}}{{\cos }^2{\mathrm{xsin}}^2\mathrm{x}}=\underset{\mathrm{x}\to 0}{lim}\frac{1}{{\cos }^2\mathrm{x}}\\ & =\frac{1}{lim(\cos \mathrm{x}{)}^2}=\frac{1}{1^2}=\frac{1}{1}=1\end{array}$

$\underset{\mathrm{x}\to 0}{lim}\frac{\frac{{\tan }^2\mathrm{x}}{{\mathrm{x}}^2}}{\frac{{\sin }^2\mathrm{x}}{{\mathrm{x}}^2}}=\underset{\mathrm{x}\to 0}{lim}\frac{\frac{\tan \mathrm{x}}{\mathrm{x}}\cdot \frac{\tan \mathrm{x}}{\mathrm{x}}}{\frac{\sin \mathrm{x}}{\mathrm{x}}\cdot \frac{\sin \mathrm{x}}{\mathrm{x}}}=\frac{\underset{\mathrm{x}\to 0}{lim}\frac{\tan \mathrm{x}}{\mathrm{x}}\cdot \frac{\tan \mathrm{x}}{\mathrm{x}}}{\underset{\mathrm{x}\to 0}{lim}\frac{\sin \mathrm{x}}{\mathrm{x}}\cdot \frac{\sin \mathrm{x}}{\mathrm{x}}}$

$=\frac{\underset{\mathrm{x}\to 0}{lim}\frac{\tan \mathrm{x}}{\mathrm{x}}\cdot \underset{\mathrm{x}\to 0}{lim}\frac{\tan \mathrm{x}}{\mathrm{x}}}{\underset{\mathrm{x}\to 0}{lim}\frac{\sin \mathrm{x}}{\mathrm{x}}\cdot \underset{\mathrm{x}\to 0}{lim}\frac{\sin \mathrm{x}}{\mathrm{x}}}=\frac{1\times 1}{1\times 1}=1$

ب- $\underset{x\to 0}{lim}[\sin 2x+\frac{\tan 4x}{6x}]$

$\begin{array}{rl}& \underset{\mathrm{x}\to 0}{lim}(2\sin \mathrm{xcos}\mathrm{x}+\frac{\tan 4\mathrm{x}}{6\mathrm{x}})\\ & =2\underset{\mathrm{x}\to 0}{lim}\sin \mathrm{x}\cdot \underset{\mathrm{x}\to 0}{lim}\cos \mathrm{x}+\underset{\mathrm{x}\to 0}{lim}\frac{\tan 4\mathrm{x}}{6\mathrm{x}}\\ & =2\underset{\mathrm{x}\to 0}{lim}\sin \mathrm{x}\cdot \underset{\mathrm{x}\to 0}{lim}\cos \mathrm{x}+\frac{4}{6}lim\frac{\tan 4\mathrm{x}}{4\mathrm{x}}\\ & =2\times 0\times 1+\frac{4}{6}\times 1=\frac{4}{6}\end{array}$

ج- $\underset{\mathrm{x}\to 0}{lim}\frac{1-\cos \mathrm{x}}{{\mathrm{x}}^2}$

$$\begin{gathered} \underset{x\to 0}{lim}\frac{1-\cos x}{x^2}=\underset{x\to 0}{lim}\frac{1-\cos x}{x^2}\cdot \frac{1+\cos x}{1+\cos x}=\underset{x\to 0}{lim}\frac{1-{\cos }^{2}x}{x^2(1+\cos x)} \\ \begin{array}{rl}& =\underset{x\to 0}{lim}\frac{{\sin }^{2}x}{x^2(1+\cos x)}=\frac{\underset{x\to 0}{lim}\frac{{\sin }^{2}x}{x^2}}{\underset{x\to 0}{lim}\frac{x^2(1+\cos x)}{x^2}}\\ & =\frac{{\left(\underset{x\to 0}{lim}\frac{\sin x}{x}\right)}^2}{\underset{x\to 0}{lim}(1+\cos x)}=\frac{1^2}{1+1}=\frac{1}{2}\end{array} \end{gathered}$$

د- $\underset{\mathrm{x}\to 0}{lim}[\frac{3\mathrm{x}}{\sin 2\mathrm{x}}+\frac{1-\cos 6\mathrm{x}}{{\sin }^2\mathrm{x}}]$

$\begin{array}{rl}1-\cos 6x=& 1-(1-2{\sin }^{2}3x)\\ & =1-1+2{\sin }^{2}3x\\ & =2{\sin }^{2}3x\\ & =\underset{x\to 0}{lim}\frac{3x}{\sin 2x}+\underset{x\to 0}{lim}\frac{2{\sin }^{2}3x}{{\sin }^{2}x}\\ & =\frac{3}{2}\underset{x\to 0}{lim}\frac{2x}{\sin 2x}+2\underset{x\to 0}{lim}\frac{{\sin }^{2}3x}{{\sin }^{2}x}\\ & =\frac{3}{2}\underset{x\to 0}{lim}\frac{1}{\sin 2x}+2\underset{x\to 0}{lim}\frac{\frac{{\sin }^{2}3x}{x^2}}{\frac{{\sin }^{2}x}{x^2}}\end{array}$

$\begin{array}{rl}& =\frac{3}{2}\times \frac{1}{1}+9\cdot 2\underset{x\to 0}{lim}\frac{\frac{{\sin }^{2}3x}{9x^2}}{\frac{{\sin }^{2}x}{x^2}}\\ & =\frac{3}{2}+18\underset{x\to 0}{lim}\frac{{\left[\frac{\sin 3x}{3x}\right]}^2}{{\left[\frac{\sin x}{x}\right]}^2}=\frac{3}{2}+18\cdot \frac{1}{1}=\frac{3}{2}+18\\ & =\frac{3+36}{2}=\frac{39}{2}\end{array}$