جد قيمة كل مما يأتي:

${\mathrm{Lim}}_{x\to 1}\frac{x^4-1}{x-1}$

$\begin{array}{rl}{\mathrm{Lim}}_{x\to 1}\frac{x^4-1}{x-1}& ={\mathrm{Lim}}_{x\to 1}\frac{(x^2-1)(x^2+1)}{x-1}={\mathrm{Lim}}_{x\to 1}\frac{(x^{-}-1)(x+1)(x^2+1)}{x-1}\\ & =(1+1)(1+1)=\left(2\right)\left(2\right)=4\end{array}$

تمارين (1-2)

1- جد قيمة كل مما يأتي:

- ${\mathrm{Lim}}_{x\to -1}(x^3+2x+3)$

${\mathrm{Lim}}_{x\to -1}(x^3+2x+3)=(-1{)}^3+2(-1)+3=-1-2+3=0$

- ${\mathrm{Lim}}_{x\to 0}\frac{x^4+1}{x+1}$

${\mathrm{Lim}}_{x\to 0}\frac{x^4+1}{x+1}=\frac{0+1}{0+1}=1$

- ${\mathrm{Lim}}_{x\to -2}\frac{x^2+2x}{x^2-x-6}$

${\mathrm{Lim}}_{x\to -2}\frac{x^2+2x}{x^2-x-6}={\mathrm{Lim}}_{x\to -2}\frac{x(x+2)}{(x-3)(x+2)}=\frac{-2}{-5}=\frac{2}{5}$

- ${\mathrm{Lim}}_{x\to 1}\frac{x^4-1}{x-1}$

$\begin{array}{rl}{\mathrm{Lim}}_{x\to 1}\frac{x^4-1}{x-1}& ={\mathrm{Lim}}_{x\to 1}\frac{(x^2-1)(x^2+1)}{x-1}={\mathrm{Lim}}_{x\to 1}\frac{(x^{-}-1)(x+1)(x^2+1)}{x-1}\\ & =(1+1)(1+1)=\left(2\right)\left(2\right)=4\end{array}$

- ${\mathrm{Lim}}_{x\to 3}\frac{x^3-27}{x^2+2x-15}$

${\mathrm{Lim}}_{x\to 3}\frac{x^3-27}{x^2+2x-15}={\mathrm{Lim}}_{x\to 3}\frac{(x-3)(x^2+3x+9)}{(x+5)(x-3)}={\mathrm{Lim}}_{x\to 3}\frac{x^2+3x+9}{x+5}=\frac{9+9+9}{8}=\frac{27}{8}$

- $Li{m}_{x\to \sqrt{2}}\frac{x^2-2}{x-\sqrt{2}}$

الطريقة الأولى:

$\begin{array}{rl}& {\mathrm{Lim}}_{x\to \sqrt{2}}\frac{(x-\sqrt{2})(x+\sqrt{2})}{x-\sqrt{2}}\\ & ={\mathrm{Lim}}_{x\to \sqrt{2}}(x+\sqrt{2})=\sqrt{2}+\sqrt{2}=2\sqrt{2}\end{array}$

الطريقة الثانية:

$={\mathrm{Lim}}_{x\to \sqrt{2}}\frac{x^2-1}{x-\sqrt{2}}\times \frac{x+\sqrt{2}}{x+\sqrt{2}}={\mathrm{Lim}}_{x\to \sqrt{2}}\frac{(x^2-2)(x+\sqrt{2})}{x^2-2}=\sqrt{2}+\sqrt{2}=2\sqrt{2}$

- ${\mathrm{Lim}}_{x\to 1}\frac{x^3+7x^2-8x}{3x^2-3}$

$\begin{array}{rl}Li{m}_{x\to 1}\frac{x^3+7x^2-8x}{3x^2-3}& =Li{m}_{x\to 1}\frac{x(x^2+7x-8)}{3(x^2-1)}={\mathrm{Lim}}_{x\to 1}\frac{x(x+8)(x-1)}{3(x-1)(x+1)}\\ & =Li{m}_{x\to 1}\frac{x(x+8)}{3(x+1)}=\frac{1(1+8)}{3(1+1)}=\frac{9}{6}=\frac{3}{2}\end{array}$

- ${\mathrm{Lim}}_{x\to -2}\frac{x^3+8}{x^4-16}$

$\begin{array}{rl}{\mathrm{Lim}}_{x\to -2}& \frac{x^3+8}{x^4-16}={\mathrm{Lim}}_{x\to -2}\frac{(x+2)(x^2-2x+4)}{(x^2-4)(x^2+4)}={\mathrm{Lim}}_{x\to -2}\frac{(x+2)(x^2-2x+4)}{(x-2)(x+2)(x^2+4)}\\ & ={\mathrm{Lim}}_{x\to -2}\frac{x^2-2x+4}{(x-2)(x^2+4)}=\frac{(-2{)}^2-2(-2)+4}{(-2-2)\left(\right(-2{)}^2+4)}=\frac{4+4+4}{(-4)\left(8\right)}=\frac{12}{-32}=\frac{-3}{8}\end{array}$

- ${\mathrm{Lim}}_{x\to 1}\frac{x^2-1}{\sqrt{x}-1}$

الطريقة الأولى:

${\mathrm{Lim}}_{x\to 1}\frac{(x-1)(x+1)}{\sqrt{x}-1}={\mathrm{Lim}}_{x\to 1}\frac{(\sqrt{x}-1)(\sqrt{x}+1)(x+1)}{\sqrt{x}-1}=(1+1)(1+1)=4$

الطريقة الثانية:

$\begin{array}{c}{\mathrm{Lim}}_{x\to 1}\frac{x^2-1}{\sqrt{x}-1}\cdot \frac{\sqrt{x}-1}{\sqrt{x}-1}={\mathrm{Lim}}_{x\to 1}\frac{(x^2-1)(\sqrt{x}+1)}{x-1}={\mathrm{Lim}}_{x\to 1}\frac{(x-1)(x+1)(\sqrt{x}+1)}{x-1}\\ =(1+1)(1+1)=4\end{array}$

- ${\mathrm{Lim}}_{x\to 3}\frac{x^2-9}{\sqrt{3x}-3}$

$\begin{array}{rl}& {\mathrm{Lim}}_{x\to 3}\frac{x^2-9}{\sqrt{3x}-3}\times \frac{\sqrt{3x}+3}{\sqrt{3x}+3}={\mathrm{Lim}}_{x\to 3}\frac{(x-3)(x+3)(\sqrt{3x}+3)}{3x-9}\\ & ={\mathrm{Lim}}_{x\to 3}\frac{(x-3)(x+3)(\sqrt{3x}+3)}{3(x-3)}=\frac{(3+3)(\sqrt{9}+3)}{3}=\frac{6\left(6\right)}{3}=12\end{array}$

- ${\mathrm{Lim}}_{x\to -1}\frac{x^2+x}{\sqrt{x+10}-3}$

$\begin{array}{rl}& {\mathrm{Lim}}_{x\to -1}\frac{x^2+x}{\sqrt{x+10}-3}=\frac{\sqrt{x+10}+3}{\sqrt{x+10}+3}\\ & {\mathrm{Lim}}_{x\to -1}\frac{x(x+1)(\sqrt{x+10}+3)}{x+10-9}={\mathrm{Lim}}_{x\to -1}\frac{x(x+1)(\sqrt{x+10}+3)}{x+1}=(-1)(\sqrt{-1+10}+3)\\ & =(-1)(3+3)=-6\end{array}$

2- إذا كانت ${\mathrm{Lim}}_{x\to 4}\frac{x^2-2x+6}{x+3}=3a-4$ جد قيمة $a\in R$؟

$\frac{(4{)}^2-2(4)+6}{2}=3a-4\Rightarrow \frac{16-8+6}{7}=3a-4\Rightarrow \frac{14}{7}=3a-4\Rightarrow 2=3a-4\Rightarrow 3a=6\Rightarrow a=2$

3- إذا كانت ${\mathrm{Lim}}_{x\to a}\frac{x^2-a^2}{x-a}=8$ جد قيمة $a\in R$؟

${\mathrm{Lim}}_{x\to a}\frac{(x-a)(x+a)}{x-a}=8\Rightarrow a+a+8\Rightarrow 2a=8\Rightarrow a=4$

4- إذا كانت $f\left(x\right)=a{x}^2+bx$ وكانت ${\mathrm{Lim}}_{x\to -2}f\left(x\right)=8 , {\mathrm{Lim}}_{x\to 1}f\left(x\right)=5$ جد قيميتي a,b الحقيقيتين

$\begin{array}{rl}& {\mathrm{Lim}}_{x\to 1}a{x}^2+bx=5\Rightarrow a(1{)}^2+b(1)=5\\ & \begin{array}{c}a+b=5\\ {\mathrm{Lim}}_{x\to -2}a{x}^2+bx=8\Rightarrow a(-2{)}^2+b(-2)=8\\ 4a-2b=8]÷2\Rightarrow 2a-b=4\end{array}\end{array}$

نحصل على (2) و(1) من:

$\begin{array}{c}a+b=5\\ 2a=b=4 \mathrm{بالجمع}\\ 3a=9\Rightarrow a=3\end{array}$

نعوض قيمة a في (1):

$3+b=5\Rightarrow b=2$

5- لتكن $f\left(x\right)=\{\begin{array}{l}{x}^2-3,x>2\\ 2-2x,x<2\end{array}$

- هل للدالة f غاية عند 2؟ بين ذلك.

$\begin{array}{rl}& {\mathrm{Lim}}_{x\to 2}{x}^2-3=(2{)}^2-3=4-3=1=L_1\\ & {\mathrm{Lim}}_{x\to 2}2-2x=2-2\left(2\right)=2-4=-2=L_2\\ & \therefore L_1\ne L_2\end{array}$

لا توجد غاية عند x=2

- جد ${\mathrm{Lim}}_{x\to 1}f\left(x\right)$

${\mathrm{Lim}}_{x\to 1}2-2x=2-2\left(1\right)=0$

6- لتكن $f\left(x\right)=\{\begin{array}{l}{x}^2+1,x\ge 2\\ 2-x,x<2\end{array}$ هل للدالة غاية عند $x\to 2$؟ بين ذلك.

$\begin{array}{rl}& {\mathrm{Lim}}_{x\to 2}{x}^2+1=(2{)}^2+1=4+1=5=L_1\\ & {\mathrm{Lim}}_{x\to 2}2-x=2-2=0=L_2\\ & \therefore L_1\ne L_2\end{array}$

لا توجد غاية عند x=2

7- لتكن $f\left(x\right)=\{\begin{array}{ll}a+2x,& x\le -1\\ 3-x^2,& x>-1\end{array}$ وكانت ${\mathrm{Lim}}_{x\to -1}f\left(x\right)$ موجودة جد قيمة a حيث $a\in R$؟

$$\begin{gathered} \because L_1=L_2 \\ \begin{array}{rl}& \therefore {\mathrm{Lim}}_{x\to -1}3-x^2=\underset{x\to -1}{lim}a+2x\Rightarrow 3-(-1{)}^2=a+2(-1)\Rightarrow 3-1=a-2\\ & 2=a-2\Rightarrow a=2+2\Rightarrow a=4\end{array} \end{gathered}$$

8- لتكن $f\left(x\right)=\{\begin{array}{l}3x+a,x\ge 3\\ x^2-b,x<3\end{array}$ وكانت ${\mathrm{Lim}}_{x\to 3}f\left(x\right)$ موجودة وأن $f\left(\sqrt{2}\right)=5$ جد قيمة $a,b\in R$؟

$\begin{array}{rl}& \left(\sqrt{2}\right)=5⟹x^2-b=5⟹(\sqrt{2}{)}^2-b=5\\ & 2-b=5\Rightarrow b=2-5\Rightarrow b=-3\\ & \therefore L_1=L_2\\ & \therefore {\mathrm{Lim}}_{x\to 3}3x+a={\mathrm{Lim}}_{x\to 3}{x}^2+3\\ & 3\left(3\right)+a=(3{)}^2+3\\ & 9+a=9+3\Rightarrow 9+a=12\Rightarrow a=12-9\Rightarrow a=3\end{array}$