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Q9- Find 
\(\lim _{x \rightarrow\left(\frac{\pi}{2}\right)^{-}}(\sec x-\tan x) \)


\(\sin \frac{\pi}{2}=1\)
 


\(\cos \frac{\pi}{2}=0 \)


\(\sec x=\frac{1}{\cos x} \)


\(\tan x=\frac{\sin x}{\cos x}\)
 

(1) 
\(\lim _{x \rightarrow \frac{\pi}{2}^{-}}(\sec x-\tan x)=\sec \left(\frac{\pi}{2}\right)-\tan \left(\frac{\pi}{2}\right) \)


\(=\frac{1}{\cos \left(\frac{\pi}{2}\right)}-\frac{\sin \left(\frac{\pi}{2}\right)}{\cos \left(\frac{\pi}{2}\right)}=\frac{1}{0}-\frac{1}{0} \)


\(=\infty-\infty \)

(2) 
\(\lim _{x \rightarrow\left(\frac{\pi}{2}\right)^{-}}(\sec x-\tan x)=\lim_{x \rightarrow (\frac{\pi}{2})^{-}} \left(\frac{1}{\cos x}-\frac{\sin x}{\cos x}\right)=\lim _{x \rightarrow \frac{\pi}{2}^{-}}\left(\frac{1-\sin x}{\cos x}\right)\)
 


\(=\frac{1-\sin \left(\frac{\pi}{2}\right)}{\cos \left(\frac{\pi}{2}\right)}=\frac{1-1}{0}=\frac{0}{0} \)

(3) 
\(\lim _{x \rightarrow\left(\frac{\pi}{2}\right)^{-}}\left(\frac{1-\sin x}{\cos y}\right)=\lim _{x \rightarrow \left(\frac{\pi}{2}\right)^{-}} \frac{-(\cos x)}{-\sin (x)}\)
 


\(=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cos x}{\sin x}=\frac{\cos \frac{\pi}{2}}{\sin \frac{\pi}{2}} \)


\(=\frac{0}{1}=0\)
 

عدد حقيقي

Q10- Find 
\(\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x} \)


\(\ln a^{b}=b \ln (a) \)


\(\frac{d \ln (x)}{d x}=\frac{1}{x}\)
 

(1) 
\(\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}=\left(1+\frac{1}{\infty}\right)^{\infty}=(1+0)^{\infty}=1^{\infty} \)

(2) 
\(y=\left(1+\frac{1}{x}\right)^{x} \Rightarrow \ln (y)=\ln \left(1+\frac{1}{x}\right)^{(x)} \Rightarrow \ln y=x \ln \left(1+\frac{1}{x}\right) \)


\(\ln y=\frac{\ln \left(1+\frac{1}{x}\right)}{\frac{1}{x}} \)

(3) 
\(\lim _{x \rightarrow \infty}\left(\frac{\ln \left(1+\frac{1}{x}\right)}{\frac{1}{x}}\right)=\frac{\ln \left(1+\frac{1}{\infty}\right)}{\frac{1}{\infty}}=\frac{\ln (1+0)}{\frac{1}{\infty}}=\frac{0}{0}\)
 

(4) 
\(\lim _{x \rightarrow \infty}\left(\frac{\ln \left(1+\frac{1}{x}\right)}{\frac{1}{x}}\right)=\lim _{x \rightarrow \infty} \frac{\frac{-\frac{1}{x^2}}{1+\frac{1}{x}}}{-\frac{1}{x^{2}}}=\lim _{x \rightarrow \infty} 1+\frac{1}{x}\)
 


\(=1+\frac{1}{\infty}=1+0=1 \)


\(\ln y=1\)
 


\(e^{\ln y}=e^1 \)


\(\Rightarrow y=e=2 \cdot 718 \)

Q11- Find 
\(\lim _{x \rightarrow \infty} \frac{x^{2}}{2^{x}} \)

(1) 
\(\lim _{x \rightarrow \infty} \frac{x^{2}}{2^{x}}=\frac{\infty^{2}}{2^{\infty}}=\frac{\infty}{\infty}\)
 


\(\lim _{x \rightarrow \infty} \frac{x^{2}}{\left(2^{x})\right.}=\lim _{x \rightarrow \infty} \frac{2 x}{2^{x} \cdot \ln (2)}=\frac{2(\infty)}{2^{\infty} \cdot \ln (2)}=\frac{\infty}{\infty}\)
 


\(\lim _{x \rightarrow \infty} \frac{x^{2}}{2^{x}}=\lim _{x \rightarrow \infty} \frac{2 x}{2^{x} \cdot \ln (2)}=\lim _{x \rightarrow \infty} \frac{2}{2^{x} \cdot \ln(2))^{2}}\)
 


\(=\frac{2}{2^{\infty} \cdot(\ln (2))^{2}} \)


\(\stackrel{\infty}{a}=\infty\)
 


\(=\frac{2}{\infty \cdot(\ln (2))^{2}}=\frac{2}{\infty}=0\)
 

Q12- Find
\(\lim _{x \rightarrow \infty} \tanh (x)\)
 


\(\tanh (x)=\frac{\sinh (x)}{\cosh (x)} \)


\(\frac{d}{d x} \cosh (x)+\sinh (x) \)


\(\frac{d}{d x} \sinh (x)=\cosh (x) \)


\(\cosh (x)=\frac{e^{x}+e^{-x}}{2} \)


\(\sinh (x)=\frac{e^{x}-e^{-x}}{2} \)

(1) 
\(\lim _{x \rightarrow \infty} \tanh (x)=\tanh (\infty)=\frac{\sinh (\infty)}{\cosh (\infty)}=\frac{\infty}{\infty} \)

(2) 
\(\lim _{x \rightarrow \infty} \frac{\sinh (x)}{\cosh (x)}=\lim _{x \rightarrow \infty} \frac{\cosh (x)}{\sinh (x)}=\frac{\infty}{\infty} \)

(3) 
\(\lim _{x \rightarrow \infty} \frac{\cosh (x)}{\sinh (x)}=\lim _{x \rightarrow \infty} \frac{\frac{\frac{x}{2}+e^{-x}}{2}}{\frac{e^{x}-e^{-x}}{2}} \)


\(\lim _{x \rightarrow \infty} \frac{e^{x}+e^{-x}}{e^{x}-e^{-x}} \)


\(\lim _{x \rightarrow \infty} \frac{e^{x}+\frac{1}{e x}}{e^{x}-\frac{1}{e x}} \)


\(=\frac{e^{\infty}+\frac{1}{e^{\infty}}}{e^{\infty}-\frac{1}{e^{\infty}}} \)


\(=\frac{e^{\infty}+0}{e^{\infty}}=\frac{\infty}{\infty} \)

(4) 
\(\lim _{x \rightarrow \infty} \frac{e^{x}+\frac{1}{e^{x}}}{e^{x}-\frac{1}{e^{x}}}= \)


\(\lim _{x \rightarrow \infty} \frac{\left(e^{x}+\frac{1}{e^{x}}\right) e^{x}}{\left(e^{x}-\frac{1}{e^{x}}\right) e^{x}} \)


\(\lim _{x \rightarrow \infty} \frac{e^{2 x}+1}{e^{2 x}-1} \)


\(=\lim _{x \rightarrow \infty} \frac{2 e^{2 x}}{2 e^{2 x}}=1\)
 

 

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