נאמר כי lim x → a f ( x ) = L \lim_{ x \to a }f(x)=L lim x → a f ( x ) = L ε > 0 \epsilon>0 ε > 0 δ > 0 \delta>0 δ > 0 0 < ∣ x − a ∣ < δ 0<|x-a|<\delta 0 < ∣ x − a ∣ < δ ∣ f ( x ) − L ∣ < ε |f(x)-L|<\epsilon ∣ f ( x ) − L ∣ < ε
הוכיחו לפי ההגדרה כי lim x → 1 ( x 2 − 2 x + 1 ) = 0 \lim_{ x \to 1 }(x^2-2x+1)=0 lim x → 1 ( x 2 − 2 x + 1 ) = 0
הוכחה: ε > 0 \epsilon>0 ε > 0 δ = ε \delta=\sqrt{ \epsilon } δ = ε 0 < ∣ x − 1 ∣ < δ 0<|x-1|<\delta 0 < ∣ x − 1∣ < δ
∣ x 2 − 2 x + 1 − 0 ∣ = ∣ ( x − 1 ) 2 ∣ = ( x − 1 ) 2 < δ 2 = ε |x^2-2x+1-0|=|(x-1)^2|=(x-1)^2<\delta^2=\epsilon ∣ x 2 − 2 x + 1 − 0∣ = ∣ ( x − 1 ) 2 ∣ = ( x − 1 ) 2 < δ 2 = ε ∎
צ"ל:
lim x → 3 ( x 2 + 1 ) = 10 \lim_{ x \to 3 } (x^2+1)=10 x → 3 lim ( x 2 + 1 ) = 10 הוכחה: ε > 0 \epsilon>0 ε > 0 δ = − 3 + 9 + ε \delta=-3+\sqrt{ 9+\epsilon } δ = − 3 + 9 + ε 0 < ∣ x − 3 ∣ < δ 0<|x-3|<\delta 0 < ∣ x − 3∣ < δ
∣ x 2 + 1 − 10 ∣ = ∣ x 2 − 9 ∣ = ∣ x − 3 ∣ ∣ x + 3 ∣ < δ ∣ x + 3 ∣ |x^2+1-10|=|x^2-9|=|x-3||x+3|<\delta |x+3| ∣ x 2 + 1 − 10∣ = ∣ x 2 − 9∣ = ∣ x − 3∣∣ x + 3∣ < δ ∣ x + 3∣ ∣ x − 3 ∣ < δ ⟹ − δ < x − 3 < δ ⟹ − 6 − δ < 6 − δ < x + 3 < 6 + δ ⟹ ∣ x + 3 ∣ < 6 + δ \begin{align*}
& |x-3|<\delta \implies -\delta<x-3<\delta \implies -6-\delta<6-\delta<x+3<6+\delta \\
& \implies |x+3|<6+\delta
\end{align*} ∣ x − 3∣ < δ ⟹ − δ < x − 3 < δ ⟹ − 6 − δ < 6 − δ < x + 3 < 6 + δ ⟹ ∣ x + 3∣ < 6 + δ כעת, נחזור למשוואה הקודמת שלנו:
δ ∣ x + 3 ∣ < δ ( 6 + δ ) = ( − 3 + 9 + ε ) ( 3 + 9 + ε ) = ε \delta |x+3|<\delta(6+\delta)=(-3+\sqrt{ 9+\epsilon })(3+\sqrt{ 9+\epsilon })=\epsilon δ ∣ x + 3∣ < δ ( 6 + δ ) = ( − 3 + 9 + ε ) ( 3 + 9 + ε ) = ε ∎
הוכיחו לפי הגדרה:
lim x → 1 3 x + 2 2 x + 3 = 1 \lim_{ x \to 1 } \frac{3x+2}{2x+3}=1 x → 1 lim 2 x + 3 3 x + 2 = 1 הוכחה: ε > 0 \epsilon>0 ε > 0 δ = min { 1 2 , 4 ε } \delta=\min\left\{ \frac{1}{2},4\epsilon \right\} δ = min { 2 1 , 4 ε } 0 < ∣ x − 1 ∣ < δ 0<|x-1|<\delta 0 < ∣ x − 1∣ < δ
∣ 3 x + 2 2 x + 3 − 1 ∣ = ∣ 3 x + 2 − 2 x − 3 2 x + 3 ∣ = ∣ x − 1 2 x + 3 ∣ = ∣ x − 1 ∣ ∣ 2 x + 3 ∣ < δ ∣ 2 x + 3 ∣ \left| \frac{3x+2}{2x+3}-1 \right|=\left| \frac{3x+2-2x-3}{2x+3} \right|=\left| \frac{x-1}{2x+3} \right|=\frac{|x-1|}{|2x+3|}< \frac{\delta}{|2x+3|} 2 x + 3 3 x + 2 − 1 = 2 x + 3 3 x + 2 − 2 x − 3 = 2 x + 3 x − 1 = ∣2 x + 3∣ ∣ x − 1∣ < ∣2 x + 3∣ δ נקח δ ≤ 1 2 \delta\leq \frac{1}{2} δ ≤ 2 1 1 2 \frac{1}{2} 2 1
∣ x − 1 ∣ < 1 2 ⟹ − 1 2 < x − 1 < 1 2 |x-1|< \frac{1}{2} \implies -\frac{1}{2}<x-1< \frac{1}{2} ∣ x − 1∣ < 2 1 ⟹ − 2 1 < x − 1 < 2 1 − 1 < 2 x − 2 < 1 ⟹ 4 < 2 x + 3 < 6 -1<2x-2<1 \implies 4<2x+3<6 − 1 < 2 x − 2 < 1 ⟹ 4 < 2 x + 3 < 6 4 < ∣ 2 x + 3 ∣ < 6 4<|2x+3|<6 4 < ∣2 x + 3∣ < 6 כעת, נחזור לאי השוויון שלנו:
δ ∣ 2 x + 3 ∣ < δ 4 ≤ ε \frac{\delta}{|2x+3|}< \frac{\delta}{4}\leq \epsilon ∣2 x + 3∣ δ < 4 δ ≤ ε ∎
צ"ל לפי הגדרה:
lim x → 1 2 x − 1 3 x − 4 = − 1 \lim_{ x \to 1 } \frac{2x-1}{3x-4}=-1 x → 1 lim 3 x − 4 2 x − 1 = − 1 הוכחה: ε > 0 \epsilon>0 ε > 0 δ = min { 1 10 , ε 10 } \delta=\min\left\{ \frac{1}{10}, \frac{\epsilon}{10} \right\} δ = min { 10 1 , 10 ε } 0 < ∣ x − 1 ∣ < δ 0<|x-1|<\delta 0 < ∣ x − 1∣ < δ
∣ 2 x − 1 3 x − 4 + 1 ∣ = ∣ 2 x − 1 + 3 x − 4 3 x − 4 ∣ = ∣ 5 x − 5 3 x − 4 ∣ = 5 ∣ x − 1 ∣ ∣ 3 x − 4 ∣ < 5 δ ∣ 3 x − 4 ∣ \left| \frac{2x-1}{3x-4}+1 \right|=\left| \frac{2x-1+3x-4}{3x-4} \right|=\left| \frac{5x-5}{3x-4} \right|=\frac{5\left| x-1 \right|}{|3x-4|}< \frac{5\delta}{|3x-4|} 3 x − 4 2 x − 1 + 1 = 3 x − 4 2 x − 1 + 3 x − 4 = 3 x − 4 5 x − 5 = ∣3 x − 4∣ 5 ∣ x − 1 ∣ < ∣3 x − 4∣ 5 δ נקח δ ≤ 1 2 \delta\leq \frac{1}{2} δ ≤ 2 1
∣ x − 1 ∣ < 1 2 ⟹ − 1 2 < x − 1 < 1 2 ⟹ − 3 2 < 3 x − 3 < 3 2 |x-1|< \frac{1}{2} \implies -\frac{1}{2}<x-1< \frac{1}{2} \implies -\frac{3}{2}<3x-3< \frac{3}{2} ∣ x − 1∣ < 2 1 ⟹ − 2 1 < x − 1 < 2 1 ⟹ − 2 3 < 3 x − 3 < 2 3 נראה כי במצב הזה לא נוכל להגיע למה שאנו רוצים.δ ≤ 1 10 \delta\leq \frac{1}{10} δ ≤ 10 1
∣ x − 1 ∣ < 1 10 ⟹ − 1 10 < x − 1 < 1 10 ⟹ − 3 10 < 3 x − 3 < 3 10 |x-1|< \frac{1}{10} \implies -\frac{1}{10}<x-1< \frac{1}{10} \implies -\frac{3}{10}<3x-3< \frac{3}{10} ∣ x − 1∣ < 10 1 ⟹ − 10 1 < x − 1 < 10 1 ⟹ − 10 3 < 3 x − 3 < 10 3 − 1.3 = − 13 10 < 3 x − 4 < − 7 10 = − 0.7 ⟹ 0.7 < ∣ 3 x − 4 ∣ < 1.3 -1.3=-\frac{13}{10}<3x-4< -\frac{7}{10}=-0.7 \implies 0.7<|3x-4|<1.3 − 1.3 = − 10 13 < 3 x − 4 < − 10 7 = − 0.7 ⟹ 0.7 < ∣3 x − 4∣ < 1.3 כעת, נחזור לאי השוויון שלנו:
5 δ ∣ 3 x − 4 ∣ ≤ 5 δ 0.7 < 10 δ ≤ ε \frac{5\delta}{|3x-4|}\leq \frac{5\delta}{0.7}< 10\delta \leq \epsilon ∣3 x − 4∣ 5 δ ≤ 0.7 5 δ < 10 δ ≤ ε ∎
מצאו כי:
lim x → 3 x = 3 \lim_{ x \to 3 } \sqrt{ x }=\sqrt{ 3 } x → 3 lim x = 3 הוכחה: ε > 0 \epsilon>0 ε > 0 δ = min { 3 , 3 ε } \delta=\min\left\{ 3,\sqrt{ 3 }\epsilon \right\} δ = min { 3 , 3 ε } 0 < ∣ x − 3 ∣ < δ 0<|x-3|<\delta 0 < ∣ x − 3∣ < δ
∣ x − 3 ∣ = ∣ x − 3 x + 3 ∣ = ∣ x − 3 ∣ x + 3 < δ x + 3 < δ 3 ≤ ε |\sqrt{ x }-\sqrt{ 3 }|=\left| \frac{x-3}{\sqrt{ x }+\sqrt{ 3 }} \right|=\frac{|x-3|}{\sqrt{ x }+\sqrt{ 3 }}< \frac{\delta}{\sqrt{ x }+\sqrt{ 3 }}< \frac{\delta}{\sqrt{ 3 }}\leq \epsilon ∣ x − 3 ∣ = x + 3 x − 3 = x + 3 ∣ x − 3∣ < x + 3 δ < 3 δ ≤ ε נבחר δ ≤ 3 \delta\leq3 δ ≤ 3
− δ < x − 3 < δ ⟹ 3 − δ < x < 3 + δ ⟹ 0 < x < 6 -\delta<x-3<\delta \implies 3-\delta<x<3+\delta \implies 0<x<6 − δ < x − 3 < δ ⟹ 3 − δ < x < 3 + δ ⟹ 0 < x < 6 ∎
נתונות 2 פונקציות f , g f,g f , g x = a x=a x = a lim x → a f ( x ) = L \lim_{ x \to a }f(x)=L lim x → a f ( x ) = L lim x → a g ( x ) = ∞ \lim_{ x \to a }g(x)=\infty lim x → a g ( x ) = ∞ lim x → a f ( x ) g ( x ) = 0 \lim_{ x \to a } \frac{f(x)}{g(x)}=0 lim x → a g ( x ) f ( x ) = 0
הוכחה: ε > 0 \epsilon>0 ε > 0 ( 1 ) (1) ( 1 ) lim x → a f ( x ) = L \lim_{ x \to a }f(x)=L lim x → a f ( x ) = L ε 1 > 0 \epsilon_{1}>0 ε 1 > 0 δ 2 > 0 \delta_{2}>0 δ 2 > 0 0 < ∣ x − a ∣ < δ 2 0<|x-a|<\delta_{2} 0 < ∣ x − a ∣ < δ 2 ∣ f ( x ) − L ∣ < ε 1 |f(x)-L|<\epsilon_{1} ∣ f ( x ) − L ∣ < ε 1 ∣ f ( x ) ∣ − ∣ L ∣ < ∣ f ( x ) − L ∣ < ε 1 |f(x)|-|L|<|f(x)-L|<\epsilon_{1} ∣ f ( x ) ∣ − ∣ L ∣ < ∣ f ( x ) − L ∣ < ε 1 ε 1 = 1 \epsilon_{1}=1 ε 1 = 1 ∣ f ( x ) ∣ < 1 + ∣ L ∣ |f(x)|<1+|L| ∣ f ( x ) ∣ < 1 + ∣ L ∣ ( 2 ) (2) ( 2 ) lim x → a g ( x ) = ∞ \lim_{ x \to a }g(x)=\infty lim x → a g ( x ) = ∞ M > 0 M>0 M > 0 δ 1 > 0 \delta_{1}>0 δ 1 > 0 0 < ∣ x − a ∣ < δ 1 0<|x-a|<\delta_{1} 0 < ∣ x − a ∣ < δ 1 g ( x ) > M g(x)>M g ( x ) > M 1 g ( x ) < 1 M \frac{1}{g(x)}< \frac{1}{M} g ( x ) 1 < M 1 M = 1 + ∣ L ∣ ε M=\frac{1+|L|}{\epsilon} M = ε 1 + ∣ L ∣ 1 ∣ g ∣ < ε 1 + ∣ L ∣ \frac{1}{|g|}< \frac{\epsilon}{1+|L|} ∣ g ∣ 1 < 1 + ∣ L ∣ ε δ = min { δ 1 , δ 2 } \delta=\min\left\{ \delta_{1},\delta_{2} \right\} δ = min { δ 1 , δ 2 } 0 < ∣ x − a ∣ < δ 0<|x-a|<\delta 0 < ∣ x − a ∣ < δ
∣ f ( x ) g ( x ) − 0 ∣ < ∣ f ( x ) g ( x ) ∣ < ( 1 ) 1 + ∣ L ∣ ∣ g ( x ) ∣ < ( 2 ) ( 1 + ∣ L ∣ ) ( ε 1 + ∣ L ∣ ) = ε \left| \frac{f(x)}{g(x)}-0 \right|< \left| \frac{f(x)}{g(x)} \right|\underset{(1)}{<} \frac{1+|L|}{|g(x)|}\underset{(2)}{<} (1+|L|)\left( \frac{\epsilon}{1+|L|} \right)=\epsilon g ( x ) f ( x ) − 0 < g ( x ) f ( x ) ( 1 ) < ∣ g ( x ) ∣ 1 + ∣ L ∣ ( 2 ) < ( 1 + ∣ L ∣ ) ( 1 + ∣ L ∣ ε ) = ε ∎
שלילת ההגדרה
הראו לפי ההגדרה כי הגבול של f ( x ) = 2 x + 3 x + 1 f(x)=\frac{2x+3}{x+1} f ( x ) = x + 1 2 x + 3 x → 3 x\to3 x → 3
הוכחה:
∣ 2 x + 3 x + 1 − 2 ∣ = ∣ 2 x + 3 − 2 x − 2 x + 1 ∣ = 1 ∣ x + 1 ∣ \left| \frac{2x+3}{x+1}-2 \right|=\left| \frac{2x+3-2x-2}{x+1} \right|=\frac{1}{|x+1|} x + 1 2 x + 3 − 2 = x + 1 2 x + 3 − 2 x − 2 = ∣ x + 1∣ 1 נדרש כי לכל δ \delta δ ∣ x − 3 ∣ < δ |x-3|<\delta ∣ x − 3∣ < δ x = 3 + δ 2 x=3+\frac{\delta}{2} x = 3 + 2 δ
1 ∣ x + 1 ∣ = 1 3 + δ 2 + 1 = 2 8 + δ \frac{1}{|x+1|}=\frac{1}{3+\frac{\delta}{2}+1}=\frac{2}{8+\delta} ∣ x + 1∣ 1 = 3 + 2 δ + 1 1 = 8 + δ 2 עבור δ ≤ 2 \delta\leq 2 δ ≤ 2 2 8 + δ > 1 5 \frac{2}{8+\delta}> \frac{1}{5} 8 + δ 2 > 5 1 ε = 1 5 \epsilon=\frac{1}{5} ε = 5 1 δ ≤ 2 \delta\leq 2 δ ≤ 2 x = 3 + δ 2 x=3+\frac{\delta}{2} x = 3 + 2 δ ∣ x − 3 ∣ = δ 2 < δ |x-3|=\frac{\delta}{2}<\delta ∣ x − 3∣ = 2 δ < δ δ > 2 \delta>2 δ > 2 x = 3.5 x=3.5 x = 3.5
1 ∣ x + 1 ∣ = 1 4.5 ≥ 1 5 \frac{1}{|x+1|}=\frac{1}{4.5} \geq \frac{1}{5} ∣ x + 1∣ 1 = 4.5 1 ≥ 5 1 לכן, התנאי מתקיים לכל δ \delta δ ∎
חישובי גבולות
מצאו את:
lim x → 1 3 x 2 − x − 2 − 3 x 2 − x + 4 \lim_{ x \to 1 } \frac{3x^2-x-2}{-3x^2-x+4} x → 1 lim − 3 x 2 − x + 4 3 x 2 − x − 2 x 1 , 2 = 1 ± 1 + 24 6 = 1 ± 5 6 = { 1 − 2 3 x_{1,2}=\frac{1\pm \sqrt{ 1+24 }}{6}= \frac{1\pm5}{6}=\begin{cases}
1 \\
-\frac{2}{3}
\end{cases} x 1 , 2 = 6 1 ± 1 + 24 = 6 1 ± 5 = { 1 − 3 2 x 1 , 2 = 1 ± 1 + 49 − 6 = 1 ± 7 6 = { − 4 3 1 x_{1,2}=\frac{1\pm \sqrt{ 1+49 }}{-6}=\frac{1\pm 7}{6}=\begin{cases}
-\frac{4}{3} \\
1
\end{cases} x 1 , 2 = − 6 1 ± 1 + 49 = 6 1 ± 7 = { − 3 4 1 lim x → 1 3 x 2 − x − 2 − 3 x 2 − x + 4 = lim x → 1 3 ( x − 1 ) ( x + 2 3 ) − 3 ( x − 1 ) ( x + 4 3 ) = − 1 + 2 3 1 + 4 3 = − 5 7 \lim_{ x \to 1 } \frac{3x^2-x-2}{-3x^2-x+4}=\lim_{ x \to 1 } \frac{3\cancel{ (x-1) }\left( x+\frac{2}{3} \right)}{-3\cancel{(x-1)}\left( x+\frac{4}{3} \right)}=- \frac{1+\frac{2}{3}}{1+\frac{4}{3}}=-\frac{5}{7} x → 1 lim − 3 x 2 − x + 4 3 x 2 − x − 2 = x → 1 lim − 3 ( x − 1 ) ( x + 3 4 ) 3 ( x − 1 ) ( x + 3 2 ) = − 1 + 3 4 1 + 3 2 = − 7 5
מצאו את:
lim x → 0 5 + 2 x − 5 − 2 x x \lim_{ x \to 0 } \frac{\sqrt{ 5+2x }-\sqrt{ 5-2x }}{x} x → 0 lim x 5 + 2 x − 5 − 2 x lim x → 0 5 + 2 x − 5 − 2 x x = lim x → 0 5 + 2 x − ( 5 − 2 x ) x ( 5 + 2 x + 5 − 2 x ) = lim x → 0 4 x x ( 5 + 2 x + 5 − 2 x ) = 4 2 5 \lim_{ x \to 0 } \frac{\sqrt{ 5+2x }-\sqrt{ 5-2x }}{x}=\lim_{ x \to 0 } \frac{5+2x-(5-2x)}{x(\sqrt{ 5+2x }+\sqrt{ 5-2x })}=\lim_{ x \to 0 } \frac{4\cancel{ x }}{\cancel{ x }(\sqrt{ 5+2x }+\sqrt{ 5-2x })} =\frac{4}{2\sqrt{ 5 }} x → 0 lim x 5 + 2 x − 5 − 2 x = x → 0 lim x ( 5 + 2 x + 5 − 2 x ) 5 + 2 x − ( 5 − 2 x ) = x → 0 lim x ( 5 + 2 x + 5 − 2 x ) 4 x = 2 5 4
מצאו את:
lim x → ∞ ( x 2 + x − x 2 + 1 ) \lim_{ x \to \infty } (\sqrt{ x^2+x }-\sqrt{ x^2+1 }) x → ∞ lim ( x 2 + x − x 2 + 1 ) = lim x → ∞ x 2 + x − ( x 2 + 1 ) x 2 + x + x 2 + 1 = lim x → ∞ x − 1 x 2 + x + x 2 + 1 = lim x → ∞ 1 − 1 x 1 + 1 x + 1 + 1 x 2 = 1 2 =\lim_{ x \to \infty } \frac{x^2+x-(x^2+1)}{\sqrt{ x^2+x }+\sqrt{ x^2+1 }}=\lim_{ x \to \infty } \frac{x-1}{\sqrt{ x^2+x }+\sqrt{ x^2+1 }}=\lim_{ x \to \infty } \frac{1-\frac{1}{x}}{\sqrt{ 1+\frac{1}{x} }+\sqrt{ 1+\frac{1}{x^2} }}=\frac{1}{2} = x → ∞ lim x 2 + x + x 2 + 1 x 2 + x − ( x 2 + 1 ) = x → ∞ lim x 2 + x + x 2 + 1 x − 1 = x → ∞ lim 1 + x 1 + 1 + x 2 1 1 − x 1 = 2 1
מצאו את:
lim x → − ∞ ( 3 x + 9 x 2 − 7 x + 1 ) \lim_{ x \to -\infty } (3x+\sqrt{ 9x^2-7x+1 }) x → − ∞ lim ( 3 x + 9 x 2 − 7 x + 1 ) = lim x → − ∞ 9 x 2 − ( 9 x 2 − 7 x + 1 ) 3 x − 9 x 2 − 7 x + 1 = lim x → − ∞ 7 x − 1 3 x − 9 x 2 − 7 x + 1 =\lim_{ x \to -\infty } \frac{9x^2-(9x^2-7x+1)}{3x-\sqrt{ 9x^2-7x+1 }}=\lim_{ x \to -\infty } \frac{7x-1}{3x-\sqrt{ 9x^2-7x+1 }} = x → − ∞ lim 3 x − 9 x 2 − 7 x + 1 9 x 2 − ( 9 x 2 − 7 x + 1 ) = x → − ∞ lim 3 x − 9 x 2 − 7 x + 1 7 x − 1 = lim x → − ∞ − 7 + 1 x − 3 − 9 − 7 x + 1 x 2 = − 7 − 3 − 3 = 7 6 =\lim_{ x \to -\infty } \frac{-7+\frac{1}{x}}{-3-\sqrt{ 9-\frac{7}{x}+\frac{1}{x^2} }}=-\frac{7}{-3-3}=\frac{7}{6} = x → − ∞ lim − 3 − 9 − x 7 + x 2 1 − 7 + x 1 = − − 3 − 3 7 = 6 7 עוד אופציה היא:
lim x → − ∞ 7 x − 1 3 x − 9 x 2 − 7 x + 1 = lim x → − ∞ 7 x − 1 3 x − x 2 ( 9 − 7 x − 1 x 2 ) = lim x → − ∞ 7 x − 1 3 x − ∣ x ∣ 9 − 7 x + 1 x 2 \lim_{ x \to -\infty } \frac{7x-1}{3x-\sqrt{ 9x^2-7x+1 }}=\lim_{ x \to -\infty } \frac{7x-1}{3x-\sqrt{ x^2\left( 9-\frac{7}{x}-\frac{1}{x^2} \right) }}=\lim_{ x \to -\infty } \frac{7x-1}{3x-|x|\sqrt{ 9-\frac{7}{x}+\frac{1}{x^2} }} x → − ∞ lim 3 x − 9 x 2 − 7 x + 1 7 x − 1 = x → − ∞ lim 3 x − x 2 ( 9 − x 7 − x 2 1 ) 7 x − 1 = x → − ∞ lim 3 x − ∣ x ∣ 9 − x 7 + x 2 1 7 x − 1 = lim x → − ∞ 7 x − 1 3 x + x 9 − 7 x + 1 x 2 = lim x → − ∞ 7 − 1 x 3 + 9 − 7 x + 1 x 2 = 7 6 =\lim_{ x \to -\infty } \frac{7x-1}{3x+x\sqrt{ 9-\frac{7}{x}+\frac{1}{x^2} }}=\lim_{ x \to -\infty } \frac{7-\frac{1}{x}}{3+\sqrt{ 9-\frac{7}{x}+\frac{1}{x^2} }}= \frac{7}{6} = x → − ∞ lim 3 x + x 9 − x 7 + x 2 1 7 x − 1 = x → − ∞ lim 3 + 9 − x 7 + x 2 1 7 − x 1 = 6 7
מצאו את:
lim x → 0 ∣ x ∣ 5 [ 7 ∣ x ∣ ] \lim_{ x \to 0 } \frac{|x|}{5}\left[ \frac{7}{|x|} \right] x → 0 lim 5 ∣ x ∣ [ ∣ x ∣ 7 ] 7 ∣ x ∣ − 1 < [ 7 ∣ x ∣ ] ≤ 7 ∣ x ∣ \frac{7}{|x|}-1< \left[ \frac{7}{|x|} \right]\leq \frac{7}{|x|} ∣ x ∣ 7 − 1 < [ ∣ x ∣ 7 ] ≤ ∣ x ∣ 7 7 5 − ∣ x ∣ 5 < ∣ x ∣ 5 [ 7 ∣ x ∣ ] ≤ 7 5 \frac{7}{5}-\frac{|x|}{5}< \frac{|x|}{5}\left[ \frac{7}{|x|} \right]\leq \frac{7}{5} 5 7 − 5 ∣ x ∣ < 5 ∣ x ∣ [ ∣ x ∣ 7 ] ≤ 5 7 ולכן, לפי סנדוויץ, הגבול הוא 7 5 \frac{7}{5} 5 7
מצאו את:
lim x → − 1 ( 2 + x ) sin ( 7 x + 1 ) \lim_{ x \to -1 } (2+x)^{\sin (\frac{7}{x+1})} x → − 1 lim ( 2 + x ) s i n ( x + 1 7 ) ( 2 + x ) sin ( 7 x + 1 ) = e ln ( 2 + x ) sin ( 7 x + 1 ) = e sin ( 7 x + 1 ) ln ( 2 x + 1 ) (2+x)^{\sin (\frac{7}{x+1})} =e^{\ln(2+x)^{\sin\left(\frac{7}{x+1}\right)}}=e^{\sin(\frac{7}{x+1})\ln(2x+1)} ( 2 + x ) s i n ( x + 1 7 ) = e l n ( 2 + x ) s i n ( x + 1 7 ) = e s i n ( x + 1 7 ) l n ( 2 x + 1 ) אנו יודעים כי lim x → − 1 sin ( 7 x + 1 ) ln ( 2 x + 1 ) = 0 \lim_{ x \to -1 }\sin\left( \frac{7}{x+1} \right)\ln(2x+1)=0 lim x → − 1 sin ( x + 1 7 ) ln ( 2 x + 1 ) = 0
lim n → ∞ e sin ( 7 x + 1 ) ln ( 2 x + 1 ) = e 0 = 1 \lim_{ n \to \infty } e^{\sin(\frac{7}{x+1})\ln(2x+1)}=e^0=1 n → ∞ lim e s i n ( x + 1 7 ) l n ( 2 x + 1 ) = e 0 = 1 הגבול המיוחד
sin x x ⟶ x → 0 1 \frac{\sin x}{x} \underset{x\to0}{\longrightarrow} 1 x sin x x → 0 ⟶ 1 אנו יודעים כי:
lim f ( x ) → 0 sin f ( x ) f ( x ) = 1 \lim_{ f(x) \to 0 } \frac{\sin f(x)}{f(x)}=1 f ( x ) → 0 lim f ( x ) sin f ( x ) = 1 לפי גבול מיוחד וגבול של הרכבה.
sin ( x − 7 ) x − 7 ⟶ x → 7 1 \frac{\sin(x-7)}{x-7}\underset{x\to7}{\longrightarrow} 1 x − 7 sin ( x − 7 ) x → 7 ⟶ 1
מצאו את:
lim x → 0 1 − cos x 3 x 2 \lim_{ x \to 0 } \frac{1-\cos x}{3x^2} x → 0 lim 3 x 2 1 − cos x = lim x → 0 2 sin 2 x 2 3 x 2 = lim x → 0 2 sin 2 x 2 3 ( x 2 ) 2 ⋅ 4 = lim x → 0 1 6 ( sin x 2 x 2 ) 2 =\lim_{ x \to 0 } \frac{2\sin^2 \frac{x}{2}}{3x^2}=\lim_{ x \to 0 } \frac{2\sin^2 \frac{x}{2}}{3 \left( \frac{x}{2} \right)^2 \cdot 4}=\lim_{ x \to 0 } \frac{1}{6}\left( \frac{\sin \frac{x}{2}}{\frac{x}{2}} \right)^2 = x → 0 lim 3 x 2 2 sin 2 2 x = x → 0 lim 3 ( 2 x ) 2 ⋅ 4 2 sin 2 2 x = x → 0 lim 6 1 ( 2 x sin 2 x ) 2 לפי sin x 2 x 2 ⟶ x → 0 1 \frac{\sin \frac{x}{2}}{\frac{x}{2}}\underset{x\to0}{\longrightarrow}1 2 x s i n 2 x x → 0 ⟶ 1
מצאו את:
lim x → 0 sin 5 x − sin 3 x sin 4 x \lim_{ x \to 0 } \frac{\sin 5x-\sin 3x}{\sin 4x} x → 0 lim sin 4 x sin 5 x − sin 3 x = lim x → 0 sin 5 x 5 x ⋅ 5 x − sin 3 x 3 x ⋅ 3 x sin 4 x 4 x ⋅ 4 x = lim x → 0 5 sin 5 x 5 x − 3 sin 3 x 3 x 4 sin 4 x 4 x = 5 − 3 4 = 1 2 =\lim_{ x \to 0 } \frac{\frac{\sin5x}{5x}\cdot 5x-\frac{\sin 3x}{3x}\cdot 3x}{\frac{\sin 4x}{4x}\cdot4x}=\lim_{ x \to 0 } \frac{5 \frac{\sin 5x}{5x}-3 \frac{\sin 3x}{3x}}{4 \frac{\sin4x}{4x}}= \frac{5-3}{4}=\frac{1}{2} = x → 0 lim 4 x s i n 4 x ⋅ 4 x 5 x s i n 5 x ⋅ 5 x − 3 x s i n 3 x ⋅ 3 x = x → 0 lim 4 4 x s i n 4 x 5 5 x s i n 5 x − 3 3 x s i n 3 x = 4 5 − 3 = 2 1 לפי הגבול הידוע וגבול של הרכבה(לכל sin k x k x \frac{\sin kx}{kx} k x s i n k x
מצאו את:
lim x → 0 sin 2 x ln ( 1 + e 3 x 2 ) \lim_{ x \to 0 } \sin^2x\ln(1+e^{\frac{3}{x^2}}) x → 0 lim sin 2 x ln ( 1 + e x 2 3 ) = lim x → 0 sin 2 x ln ( e 3 x 2 ( e − 3 x 2 + 1 ) ) = lim x → 0 sin 2 x ( 3 x 2 + ln ( e − 3 x 2 + 1 ) ) =\lim_{ x \to 0 } \sin^2x\ln\left( e^{\frac{3}{x^2}}\left( e^{\frac{-3}{x^2}}+1 \right) \right)=\lim_{ x \to 0 } \sin^2x\left( \frac{3}{x^2}+\ln(e^{-\frac{3}{x^2}}+1) \right) = x → 0 lim sin 2 x ln ( e x 2 3 ( e x 2 − 3 + 1 ) ) = x → 0 lim sin 2 x ( x 2 3 + ln ( e − x 2 3 + 1 ) ) = lim x → 0 3 sin 2 x x 2 + sin 2 x ln ( e − 3 x 2 + 1 ) = 3 ⋅ 1 2 + 0 ⋅ 0 = 3 =\lim_{ x \to 0 } 3\frac{\sin^2x}{x^2}+ \sin^2 x\ln(e^{-\frac{3}{x^2}}+1)=3\cdot 1^2 + 0 \cdot 0=3 = x → 0 lim 3 x 2 sin 2 x + sin 2 x ln ( e − x 2 3 + 1 ) = 3 ⋅ 1 2 + 0 ⋅ 0 = 3
מצאו את:
lim x → 1 x sin ( x 2 − 1 ) tan ( x 3 − 1 ) \lim_{ x \to 1 } \frac{\sqrt{ x }\sin(x^2-1)}{\tan(x^3-1)} x → 1 lim tan ( x 3 − 1 ) x sin ( x 2 − 1 ) לכל x ≠ 1 x\neq1 x = 1
x sin ( x 2 − 1 ) tan ( x 3 − 1 ) = x sin ( x 2 − 1 ) sin ( x 3 − 1 ) cos ( x 3 − 1 ) \frac{\sqrt{ x }\sin(x^2-1)}{\tan(x^3-1)}=\frac{\sqrt{ x }\sin(x^2-1)}{\sin(x^3-1)}\cos(x^3-1) tan ( x 3 − 1 ) x sin ( x 2 − 1 ) = sin ( x 3 − 1 ) x sin ( x 2 − 1 ) cos ( x 3 − 1 ) x ⟶ x → 1 1 \sqrt{ x }\underset{x\to{1}}{\longrightarrow} 1 x x → 1 ⟶ 1 cos ( x 3 − 1 ) ⟶ x → 1 1 \cos(x^3-1)\underset{x\to1}{\longrightarrow}1 cos ( x 3 − 1 ) x → 1 ⟶ 1 כי x 3 − 1 → 0 x^3-1\to0 x 3 − 1 → 0
sin ( x 2 − 1 ) sin ( x 3 − 1 ) = sin ( x 2 − 1 ) x 2 − 1 ⋅ x 3 − 1 sin ( x 3 − 1 ) ⋅ x 2 − 1 x 3 − 1 ⟶ x → 1 1 ⋅ 1 ⋅ 2 3 \frac{\sin(x^2-1)}{\sin(x^3-1)}=\frac{\sin(x^2-1)}{x^2-1}\cdot \frac{x^3-1}{\sin(x^3-1)}\cdot \frac{x^2-1}{x^3-1} \underset{x\to 1}{\longrightarrow} 1\cdot 1\cdot \frac{2}{3} sin ( x 3 − 1 ) sin ( x 2 − 1 ) = x 2 − 1 sin ( x 2 − 1 ) ⋅ sin ( x 3 − 1 ) x 3 − 1 ⋅ x 3 − 1 x 2 − 1 x → 1 ⟶ 1 ⋅ 1 ⋅ 3 2 x 2 − 1 x 3 − 1 = ( x − 1 ) ( x + 1 ) ( x − 1 ) ( x 2 + x + 1 ) ⟶ x → 1 2 3 \frac{x^2-1}{x^3-1}=\frac{\cancel{ (x-1) }(x+1)}{\cancel{(x-1)}(x^2+x+1)}\underset{x\to 1}{\longrightarrow} \frac{2}{3} x 3 − 1 x 2 − 1 = ( x − 1 ) ( x 2 + x + 1 ) ( x − 1 ) ( x + 1 ) x → 1 ⟶ 3 2 x sin ( x 2 − 1 ) sin ( x 3 − 1 ) cos ( x 3 − 1 ) ⟶ x → 1 1 ⋅ 2 3 ⋅ 1 = 2 3 \frac{\sqrt{ x }\sin(x^2-1)}{\sin(x^3-1)}\cos(x^3-1)\underset{x\to1}{\longrightarrow} 1\cdot \frac{2}{3}\cdot 1=\frac{2}{3} sin ( x 3 − 1 ) x sin ( x 2 − 1 ) cos ( x 3 − 1 ) x → 1 ⟶ 1 ⋅ 3 2 ⋅ 1 = 3 2