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Question Number 162834 by mkam last updated on 01/Jan/22

Answered by abdullahhhhh last updated on 01/Jan/22

lim_(x→0) ((x+tan2x)/(x−tan2x)) /x lim_(x→0) ((1+2)/(1−2))=−3

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{tan}}\mathrm{2}\boldsymbol{\mathrm{x}}}{\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{tan}}\mathrm{2}\boldsymbol{\mathrm{x}}}\:/\boldsymbol{\mathrm{x}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}+\mathrm{2}}{\mathrm{1}−\mathrm{2}}=−\mathrm{3} \\ $$$$ \\ $$

Answered by abdullahhhhh last updated on 01/Jan/22

lim_(x→e) ((sinx−x)/x^3 )=((sin(e)−e)/e^3 )

$$\underset{{x}\rightarrow\boldsymbol{\mathrm{e}}} {\mathrm{lim}}\:\frac{\boldsymbol{\mathrm{sinx}}−\boldsymbol{\mathrm{x}}}{\boldsymbol{\mathrm{x}}^{\mathrm{3}} }=\frac{\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{e}}\right)−\boldsymbol{\mathrm{e}}}{\boldsymbol{\mathrm{e}}^{\mathrm{3}} } \\ $$$$ \\ $$

Answered by abdullahhhhh last updated on 01/Jan/22

lim_(x→0) ((e^x +e^(−x) −x^2 −2)/(sin^2 x−x^2 ))=((0/0)) use lobital lim_(x→0) ((e^x −e^(−x) −2x)/(2sinx cosx(sin2x)−2x))=(0/0) lim_(x→0) ((e^x +e^(−x) −2)/(2cos2x−2))=((2−2)/(2−2))=(0/0) lim_(x→0) ((e^x −e^(−x) )/(−4sin2x))=((1−1)/0)=(0/0) lim_(x→0) ((e^x +e^(−x) )/(−8cos2x))=(2/(−8))=((−1)/4)

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{x}}} +\boldsymbol{\mathrm{e}}^{−\boldsymbol{\mathrm{x}}} −\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{2}}{\boldsymbol{\mathrm{sin}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{x}}^{\mathrm{2}} }=\left(\frac{\mathrm{0}}{\mathrm{0}}\right)\:\boldsymbol{\mathrm{use}}\:\boldsymbol{\mathrm{lobital}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{x}}} −\boldsymbol{\mathrm{e}}^{−\boldsymbol{\mathrm{x}}} −\mathrm{2}\boldsymbol{\mathrm{x}}}{\mathrm{2}\boldsymbol{\mathrm{sinx}}\:\boldsymbol{\mathrm{cosx}}\left(\boldsymbol{\mathrm{sin}}\mathrm{2}\boldsymbol{\mathrm{x}}\right)−\mathrm{2}\boldsymbol{\mathrm{x}}}=\frac{\mathrm{0}}{\mathrm{0}}\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{x}}} +\boldsymbol{\mathrm{e}}^{−\boldsymbol{\mathrm{x}}} −\mathrm{2}}{\mathrm{2}\boldsymbol{\mathrm{cos}}\mathrm{2}\boldsymbol{\mathrm{x}}−\mathrm{2}}=\frac{\mathrm{2}−\mathrm{2}}{\mathrm{2}−\mathrm{2}}=\frac{\mathrm{0}}{\mathrm{0}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{x}}} −\boldsymbol{\mathrm{e}}^{−\boldsymbol{\mathrm{x}}} }{−\mathrm{4}\boldsymbol{\mathrm{sin}}\mathrm{2}\boldsymbol{\mathrm{x}}}=\frac{\mathrm{1}−\mathrm{1}}{\mathrm{0}}=\frac{\mathrm{0}}{\mathrm{0}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{x}}} +\boldsymbol{\mathrm{e}}^{−\boldsymbol{\mathrm{x}}} }{−\mathrm{8}\boldsymbol{\mathrm{cos}}\mathrm{2}\boldsymbol{\mathrm{x}}}=\frac{\mathrm{2}}{−\mathrm{8}}=\frac{−\mathrm{1}}{\mathrm{4}} \\ $$

Answered by abdullahhhhh last updated on 01/Jan/22

lim_(x→0) ((xe^x −ln(1+x))/x^2 )=(0/0) lobital lim_(x→0) ((xe^x +e^x −(1/((x+1) )))/(2x))=(0/0) lim_(x→0) ((xe^x +e^x +e^x +(1/((x+1)^2 )))/2)=(3/2)

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\boldsymbol{\mathrm{xe}}^{\boldsymbol{\mathrm{x}}} −\boldsymbol{\mathrm{ln}}\left(\mathrm{1}+\boldsymbol{\mathrm{x}}\right)}{\boldsymbol{\mathrm{x}}^{\mathrm{2}} }=\frac{\mathrm{0}}{\mathrm{0}}\:\boldsymbol{\mathrm{lobital}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\boldsymbol{\mathrm{xe}}^{\boldsymbol{\mathrm{x}}} +\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{x}}} −\frac{\mathrm{1}}{\left(\boldsymbol{\mathrm{x}}+\mathrm{1}\right)\:}}{\mathrm{2}\boldsymbol{\mathrm{x}}}=\frac{\mathrm{0}}{\mathrm{0}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\boldsymbol{\mathrm{xe}}^{\boldsymbol{\mathrm{x}}} +\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{x}}} +\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{x}}} +\frac{\mathrm{1}}{\left(\boldsymbol{\mathrm{x}}+\mathrm{1}\right)^{\mathrm{2}} }}{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$

Answered by abdullahhhhh last updated on 01/Jan/22

lim_(x→∞) (((lnx)/x))=(∞/∞) use lobital lim_(x→∞) (1/x)=(1/∞)=0

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\boldsymbol{\mathrm{lnx}}}{\boldsymbol{\mathrm{x}}}\right)=\frac{\infty}{\infty}\:\boldsymbol{\mathrm{use}}\:\boldsymbol{\mathrm{lobital}} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}=\frac{\mathrm{1}}{\infty}=\mathrm{0} \\ $$

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