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Question Number 80064 by Rio Michael last updated on 30/Jan/20

$$\underset{x\to -\mathrm{\infty }}{\lim }\left[\sqrt{1-{xe}^x}\right]$$

Commented by mathmax by abdo last updated on 30/Jan/20

$$noitsaindeterminedformitsnotcorrect$$

Commented by mr W last updated on 30/Jan/20

$$lett=-x$$$$t\to +\mathrm{\infty }whenx\to -\mathrm{\infty }$$$$\underset{x\to -\mathrm{\infty }}{\lim }(1-{xe}^x)=\underset{t\to +\mathrm{\infty }}{\lim }(1+\frac{t}{e^t})$$$$=1+\underset{t\to +\mathrm{\infty }}{\lim }\left(\frac{t}{e^t}\right)$$$$=1+\underset{t\to +\mathrm{\infty }}{\lim }\left(\frac{1}{e^t}\right)$$$$=1+0$$$$=1$$$$\Rightarrow \underset{x\to -\mathrm{\infty }}{\lim }\left[\sqrt{1-{xe}^x}\right]=\sqrt{1}=1$$

Commented by Rio Michael last updated on 30/Jan/20

$$thankssir$$

Commented by Rio Michael last updated on 30/Jan/20

$$\mathrm{sir}\mathrm{i}\mathrm{solved}\mathrm{it}\mathrm{like}\mathrm{this}:$$$$\underset{x\to -\mathrm{\infty }}{\lim }\left[\sqrt{1-{xe}^x}\right]=\sqrt{1-(-\mathrm{\infty })e^{-\mathrm{\infty }}}$$$$=\sqrt{1+\frac{+\mathrm{\infty }}{e^{\mathrm{\infty }}}}=\sqrt{1+0}=1$$$$$$

Commented by Rio Michael last updated on 30/Jan/20

$$youmeanimustdothesubstitutiont=-x?$$

Commented by mathmax by abdo last updated on 04/Feb/20

$$letf\left(x\right)=\sqrt{1-{xe}^x}cha7gement{e}^x=tgivex=ln\left(t\right)$$$$\Rightarrow {lim}_{x\to -\mathrm{\infty }}f\left(x\right)={lim}_{t\to 0^{+}}\sqrt{1-tlnt}=\sqrt{1-o}=1$$

Answered by Kamel Kamel last updated on 30/Jan/20

$$\mathrm{We}\mathrm{have}:\mathrm{\forall }\mathrm{x}⋘0,0<-{\mathrm{xe}}^{\mathrm{x}}<1$$$$\mathrm{So}:\sqrt{1-{\mathrm{xe}}^{\mathrm{x}}}=1+\alpha /\alpha <1$$$$\therefore \left[\sqrt{1-{\mathrm{xe}}^{\mathrm{x}}}\right]=1$$$$\therefore \underset{x\to -\mathrm{\infty }}{\lim }\left[\sqrt{1-{xe}^x}\right]=1$$

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