ساعدوووووووووني في حساب النهايات

[samar biologie]

ساعدوني ربي يحفظكم عااااااااجل


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[samar biologie]

................................

[مُسافر]

سأحاول حل بعضها نظرا لضيق الوقت

نبدأ بالنهاية التاسعة (نعتمد على تعريف العدد المشتق )



http://latex.codecogs.com/gif.latex?\dpi{150}%20l=\lim_{x\rightarrow%200}(x+ 1)^{\frac{1}{x}}%20\\%20\\%20\ln(l)=\lim_{x\righta rrow%200}\frac{ln(x+1)}{x}%20\\%20let:f(x)=ln(x+1) %20\\%20then:%20\\%20\ln(l)=\lim_{x\rightarrow%200 }\frac{ln(x+1)-ln(0+1)}{x-0}=f%27(0)=1%20\\%20so:%20l=e^{ln(l)}=e

النهاية الثامنة

نقوم بالتالي
http://latex.codecogs.com/gif.latex?\dpi{150}%20\lim_{x\rightarrow%20\infty% 20}(\frac{1}{x}+1)^x%20\\%20let:x=\frac{1}{y},when :x\rightarrow%20\infty,y\rightarrow%200%20\\%20\li m_{x\rightarrow%20\infty%20}(\frac{1}{x}+1)^x=\lim _{y\rightarrow%200%20}(y+1)^{\frac{1}{y}}

وننجز برهان مماثل لماسبق لنجد النتيجة هي e


النهاية السابعة

http://latex.codecogs.com/gif.latex?\dpi{150}%20\lim_{x\rightarrow%200}x^x=\ lim_{x\rightarrow%200}e^{x\ln(x)}%20\\%20\\%20we%2 0\%20know%20\%20that:%20\lim_{x\rightarrow%200}x\l n(x)=0%20\\%20so:%20\\%20\lim_{x\rightarrow%200}e^ {x\ln(x)}=e^0=1

[مُسافر]

النهاية الثانية عشر

http://latex.codecogs.com/gif.latex?\dpi{150}%20\lim_{x\rightarrow%20\frac{\ pi}{4}}\frac{1-tan(x)}{cos(2x)}=\lim_{x\rightarrow%20\frac{\pi}{4 }}\frac{cos(x)-sin(x)}{cos(x)(cos^2(x)-sin^2(x))}%20\\%20\\%20=\lim_{x\rightarrow%20\frac {\pi}{4}}\frac{cos(x)-sin(x)}{cos(x)(cos(x)-sin(x))(cos(x)+sin(x))}%20\\%20\\%20=\lim_{x\right arrow%20\frac{\pi}{4}}\frac{1}{cos(x)(cos(x)+sin(x ))}%20\\%20\\%20=\frac{1}{\frac{1}{\sqrt{2}}(\frac {1}{\sqrt{2}}+\frac{1}{\sqrt{2}})}=1

النهاية العاشرة
http://latex.codecogs.com/gif.latex?\dpi{150}%20\lim_{x\rightarrow%200}\frac {1}{1+e^{\frac{1}{x}}}%20\\%20first:\lim_{x\overse t{%3E}{\rightarrow}%200}\frac{1}{1+e^{\frac{1}{x}} }=0%20\\%20second:\lim_{x\overset{%3C}{\rightarrow }%200}\frac{1}{1+e^{\frac{1}{x}}}=1%20\\%20\\%20\% 20limit%20\%20does%20\%20not%20\%20exist%20\%20at% 20:%20x\rightarrow%200


النهاية الخامسة عشر

http://latex.codecogs.com/gif.latex?\dpi{150}%20\lim_{x\rightarrow%200}\frac {1-cos(x)}{sin(x)}=\lim_{x\rightarrow%200}\frac{x}{si n(x)}\times%20\frac{1-cos(x)}{x}%20\\%20\\%20=\lim_{x\rightarrow%200}-%20\frac{cos(x)-1}{x}%20\\%20\\%20let:f(x)=cos(x)%20\\%20then:%20\ \%20-1\times%20\lim_{x\rightarrow%200}%20\frac{cos(x)-1}{x}=-1\times\lim_{x\rightarrow%200}%20\frac{cos(x)-cos(0)}{x-0}=-%20f%27(0)%20=-1\times0=0

[samar biologie]

النهاية الثانية عشر

http://latex.codecogs.com/gif.latex?\dpi{150}%20\lim_{x\rightarrow%20\frac{\ pi}{4}}\frac{1-tan(x)}{cos(2x)}=\lim_{x\rightarrow%20\frac{\pi}{4 }}\frac{cos(x)-sin(x)}{cos(x)(cos^2(x)-sin^2(x))}%20\\%20\\%20=\lim_{x\rightarrow%20\frac {\pi}{4}}\frac{cos(x)-sin(x)}{cos(x)(cos(x)-sin(x))(cos(x)+sin(x))}%20\\%20\\%20=\lim_{x\right arrow%20\frac{\pi}{4}}\frac{1}{cos(x)(cos(x)+sin(x ))}%20\\%20\\%20=\frac{1}{\frac{1}{\sqrt{2}}(\frac {1}{\sqrt{2}}+\frac{1}{\sqrt{2}})}=1

النهاية العاشرة
http://latex.codecogs.com/gif.latex?\dpi{150}%20\lim_{x\rightarrow%200}\frac {1}{1+e^{\frac{1}{x}}}%20\\%20first:\lim_{x\overse t{%3E}{\rightarrow}%200}\frac{1}{1+e^{\frac{1}{x}} }=0%20\\%20second:\lim_{x\overset{%3C}{\rightarrow }%200}\frac{1}{1+e^{\frac{1}{x}}}=1%20\\%20\\%20\% 20limit%20\%20does%20\%20not%20\%20exist%20\%20at% 20:%20x\rightarrow%200


النهاية الخامسة عشر

http://latex.codecogs.com/gif.latex?\dpi{150}%20\lim_{x\rightarrow%200}\frac {1-cos(x)}{sin(x)}=\lim_{x\rightarrow%200}\frac{x}{si n(x)}\times%20\frac{1-cos(x)}{x}%20\\%20\\%20=\lim_{x\rightarrow%200}-%20\frac{cos(x)-1}{x}%20\\%20\\%20let:f(x)=cos(x)%20\\%20then:%20\ \%20-1\times%20\lim_{x\rightarrow%200}%20\frac{cos(x)-1}{x}=-1\times\lim_{x\rightarrow%200}%20\frac{cos(x)-cos(0)}{x-0}=-%20f%27(0)%20=-1\times0=0


باااارك الله فيك اخي ربي يحفظك وينجحك

لو تزيدلي 13 و 14
جزاك الله خيرا اخي :rolleyes:

[مُسافر]

النهاية الرابعة عشر :

باستخدام تعريف العدد المشتق

http://latex.codecogs.com/gif.latex?\dpi{150}%20\lim_{x\rightarrow0%20}\frac {a^x-b^x}{sin(x)}=\lim_{x\rightarrow0%20}\frac{a^x-b^x}{sin(x)}\times%20\frac{x}{x}%20\\%20\\%20=\lim _{x\rightarrow0%20}\frac{x}{sin(x)}\times%20\frac{ a^x-b^x}{x}%20\\%20\\%20=\lim_{x\rightarrow0%20}1\time s%20\frac{a^x-b^x}{x}=\lim_{x\rightarrow0%20}%20\frac{a^x-b^x}{x}%20\\%20\\%20=\lim_{x\rightarrow0%20}%20\fr ac{a^x-1}{x}-\lim_{x\rightarrow0%20}\frac{b^x-1}{x}%20\\%20let:f(x)=a^x=e^{xln(a)};and%20\%20:g( x)=b^x=e^{xln(b)}%20\\%20\\%20=%20\lim_{x\rightarr ow0%20}%20\frac{a^x-1}{x}-\lim_{x\rightarrow0%20}\frac{b^x-1}{x}%20=\lim_{x\rightarrow0%20}%20\frac{a^x-a^0}{x-0}-\lim_{x\rightarrow0%20}\frac{b^x-b^0}{x-0}=f%27(0)-g%27(0)%20\\%20\\%20=\ln(a)-\ln(b)%20\\%20\\%20=\ln(\frac{a}{b})

النهاية الثالثة عشر

http://latex.codecogs.com/gif.latex?\dpi{150}%20\lim_{x\rightarrow%20\infty% 20}\frac{x}{\sqrt{1+x^2}}=\lim_{x\rightarrow%20\in fty%20}\frac{x}{x\sqrt{\frac{1}{x^2}+1}}=\lim_{x\r ightarrow%20\infty%20}\frac{1}{\sqrt{\frac{1}{x^2} +1}}=1%20\\%20\\%20\lim_{x\rightarrow%20-\infty%20}\frac{x}{\sqrt{1+x^2}}=\lim_{x\rightarro w-%20\infty%20}\frac{x}{-x\sqrt{\frac{1}{x^2}+1}}=\lim_{x\rightarrow-%20\infty%20}\frac{-1}{\sqrt{\frac{1}{x^2}+1}}=-1

[samar biologie]

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