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Question Number 28151 by tawa tawa last updated on 21/Jan/18

$$\underset{x\to 1}{\lim }(\frac{1}{{\log }_{\mathrm{e}}\mathrm{x}}-\frac{\mathrm{x}}{\mathrm{x}-1})$$

Commented by çhëý böý last updated on 21/Jan/18

$$\underset{x\to 1}{\lim }\left(\frac{(x-1)-xlnx}{(x-1)lnx}\right)=\frac{0}{0}$$$$\underset{x\to 1}{\lim }\left(\frac{x-(\frac{x}{x}+lnx)}{lnx+\frac{(x-1)}{x}}\right)$$$$\underset{x\to 1}{\lim }\left(\frac{x(x-1+1nx)}{xlnx+(x-1)}\right)=\frac{0}{0}$$$$\underset{x\to 1}{\lim }\left(\frac{(x-1+lnx)+x(1-0+\frac{1}{x})}{1+lnx+1}\right)$$$$\underset{x\to 1}{\lim }\left(\frac{0+0+2}{1+0+1}\right)=\frac{2}{2}=1$$

Commented by tawa tawa last updated on 21/Jan/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by abdo imad last updated on 21/Jan/18

$$={lim}_{x\to 1}(\frac{1}{lnx}-\frac{x}{x-1})letusethech.x-1=t$$$$x\to 1\iff t\to olim(...)={lim}_{t\to 0}(\frac{1}{ln(1+t)}-\frac{1+t}{t})$$$$={lim}_{t\to 0}\frac{t-(1+t)ln(1+t)}{tln(1+t)}lettakef\left(t\right)=t-(1+t)ln(1+t)$$$$andg\left(t\right)=tln(1+t)wehave$$$$f^{{}^{\prime }}\left(t\right)=1-(ln(1+t)+1)=-ln(1+t)and{f}^{{}^{″}}\left(t\right)=\frac{-1}{1+t}$$$$g^{{}^{\prime }}\left(t\right)=ln(1+t)+\frac{t}{1+t}=ln(1+t)+1-\frac{1}{1+t}and$$$$g^{{}^{″}}\left(t\right)=\frac{1}{1+t}+\frac{1}{{(1+t)}^2}so$$$${lim}_{x\to 1}(\frac{1}{lnx}-\frac{x}{x-1})={lim}_{t\to 0}\frac{f^{{}^{″}}\left(t\right)}{g^{{}^{″}}\left(t\right)}=\frac{-1}{2}.$$

Commented by abdo imad last updated on 21/Jan/18

$$wehaveusedhospitaltheorem.$$

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