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Question Number 73555 by Rio Michael last updated on 13/Nov/19

$$find$$$$a)\underset{x\to -\mathrm{\infty }}{Lim}\frac{ln(1+e^x)}{e^x}$$$$b)\underset{x\to +\mathrm{\infty }}{lim}(\sqrt{x^2+3x}-x)$$$$c)\underset{x\to 0}{lim}\sqrt{x}ln\left(sinx\right)$$$$d)\underset{x\to +\mathrm{\infty }}{lim}\frac{e^{2x+1}-e^x}{x^2-x-1}$$$$e)\underset{x\to -\mathrm{\infty }}{lim}\left[\sqrt{1-{xe}^x}\right]$$

Commented by mathmax by abdo last updated on 13/Nov/19

$$a)weusechang.e^x=t\Rightarrow {lim}_{x\to -\mathrm{\infty }}\frac{ln(1+e^x)}{e^x}={lim}_{t\to 0}\frac{ln(1+t)}{t}$$$$=1$$$$b)wehave\sqrt{x^2+3x}-x=\sqrt{x^2(1+\frac{3}{x})}-x$$$$=x\sqrt{1+\frac{3}{x}}-x\sim x(1+\frac{3}{2x})-x(x\to +\mathrm{\infty })$$$$\sim \frac{3}{2}\Rightarrow {lim}_{x\to +\mathrm{\infty }}\sqrt{x^2+3x}-x=\frac{3}{2}$$$$c){lim}_{x\to 0}\sqrt{x}ln\left(sinx\right)?wehavesinx\sim x\Rightarrow \sqrt{x}ln\left(sinx\right)\sim \sqrt{x}ln\left(x\right)$$$$and{lim}_{x\to 0}\sqrt{x}ln\left(x\right){=}_{\sqrt{x}=u}{lim}_{u\to 0}uln\left(u^2\right)={lim}_{u\to 0}2uln\left(u\right)=0$$$$$$

Commented by mathmax by abdo last updated on 13/Nov/19

$$d)letf\left(x\right)=\frac{e^{2x+1}-e^x}{x^2-x-1}\Rightarrow f\left(x\right)=\frac{e^{2x+1}}{x^2}\times \frac{1-e^{x-2x-1}}{1-\frac{1}{x}-\frac{1}{x^2}}$$$$wehave{lim}_{x\to +\mathrm{\infty }}\frac{e^{2x+1}}{x^2}=+\mathrm{\infty }(e^{x}defeat{x}^n...)$$$${lim}_{x\to +\mathrm{\infty }}\frac{1-e^{-x-1}}{1-\frac{1}{x}-\frac{1}{x^2}}=1\Rightarrow {lim}_{x\to +\mathrm{\infty }}f\left(x\right)=+\mathrm{\infty }$$

Commented by Rio Michael last updated on 14/Nov/19

$$thankssir$$

Commented by malwaan last updated on 14/Nov/19

$$\mathit{b})\underset{\mathit{x}\to \mathrm{\infty }}{\mathit{l}\mathit{i}\mathit{m}}\frac{(\sqrt{{\mathit{x}}^2+3\mathit{x}}-\mathit{x})(\sqrt{{\mathit{x}}^2+3\mathit{x}}+\mathit{x})}{\sqrt{{\mathit{x}}^2+3\mathit{x}}+\mathit{x}}$$$$=\underset{\mathit{x}\to \mathrm{\infty }}{\mathit{l}\mathit{i}\mathit{m}}\frac{{\mathit{x}}^2+3\mathit{x}-{\mathit{x}}^2}{\mathit{x}[\sqrt{1+\frac{3}{\mathit{x}}}+1]}$$$$=\underset{\mathit{x}\to \mathrm{\infty }}{\mathit{l}\mathit{i}\mathit{m}}\frac{3}{\sqrt{1+\frac{3}{\mathit{x}}}+1}=\frac{3}{2}$$

Answered by ajfour last updated on 13/Nov/19

$$a)e^{-\mathrm{\infty }}\to 0and\underset{h\to 0}{\lim }\frac{\ln (1+h)}{h}=1$$$$Ans.to\left(a\right)=1$$$$b)\underset{x\to +\mathrm{\infty }}{lim}(\sqrt{x^2+3x}-x)$$$$=\underset{x\to \mathrm{\infty }}{\lim }\frac{3x}{\sqrt{x^2+3x}+x}=\underset{x\to \mathrm{\infty }}{\lim }\frac{3}{\sqrt{1+\frac{3}{x}}+1}$$$$=\frac{3}{2}$$$$c)\underset{x\to 0}{lim}\sqrt{x}ln\left(sinx\right)$$$$=\frac{\underset{x\to 0}{\lim }\frac{d}{dx}\ln \left(\sin x\right)}{\underset{x\to 0}{\lim }\frac{d}{dx}\left(\frac{1}{\sqrt{x}}\right)}=\frac{\underset{x\to 0}{\lim }\frac{1}{\tan x}}{\underset{x\to 0}{\lim }(-\frac{1}{2x\sqrt{x}})}$$$$=\underset{x\to 0}{\lim }\left(\frac{x}{\tan x}\right)\times \underset{x\to 0}{\lim }(-2\sqrt{x})$$$$=1\times 0=0.$$$$d)\underset{x\to +\mathrm{\infty }}{lim}\frac{e^{2x+1}-e^x}{x^2-x-1}$$$$=\underset{x\to \mathrm{\infty }}{\lim }{e}^x\left(\frac{e\times e^x-1}{x^2-x+1}\right)$$$$=\mathrm{\infty }.$$

Commented by Rio Michael last updated on 13/Nov/19

$$thankssir$$

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