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Question Number 173765 by saly last updated on 17/Jul/22

Answered by cortano1 last updated on 18/Jul/22

$$L=\underset{x\to 1}{\lim }(\frac{3}{1-\sqrt{x}}-\frac{2}{1-\sqrt[3]{x}})=?$$$$[letx=t^6\wedge t\to 1]$$$$L=\underset{t\to 1}{\lim }(\frac{3}{1-t^3}-\frac{2}{1-t^2})$$$$=\underset{t\to 1}{\lim }(\frac{3}{(1-t)(1+t+t^2)}-\frac{2}{(1-t)(1+t)})$$$$=\underset{t\to 1}{\lim }\left(\frac{3t+3-2t^2-2t-2}{(1-t)(1+t)(1+t+t^2)}\right)$$$$=\frac{1}{6}\times \underset{t\to 1}{\lim }\left(\frac{-(t-1)(2t+1)}{-(t-1)}\right)$$$$=\frac{1}{6}\times 3=\frac{1}{2}$$

Commented by saly last updated on 18/Jul/22

$$Thankyou$$

Answered by blackmamba last updated on 18/Jul/22

$$setx-1=r\wedge r\to 0$$$$=\underset{r\to 0}{\lim }(\frac{3}{1-\sqrt{1+r}}-\frac{2}{1-\sqrt[3]{1+r}})$$$$=\underset{r\to 0}{\lim }\left(\frac{3(1-\sqrt[3]{1+r})-2(1-\sqrt{1+r})}{(1-\sqrt{1+r})(1-\sqrt[3]{1+r})}\right)$$$$=\underset{r\to 0}{\lim }(\frac{3(1-(1+\frac{r}{3}-\frac{1}{9}{r}^2))-2(1-(1+\frac{1}{2}r-\frac{1}{8}{r}^2))}{(1-(1+\frac{1}{2}r))(1-(1+\frac{1}{3}r))}$$$$=\underset{r\to 0}{\lim }\left(\frac{3(-\frac{r}{3}+\frac{r^2}{9})-2(-\frac{r}{2}+\frac{r^2}{8})}{(-\frac{1}{2}r)(-\frac{1}{3}r)}\right)$$$$=\underset{r\to 0}{\lim }\left(\frac{\frac{r^2}{3}-\frac{r^2}{4}}{\frac{1}{6}{r}^2}\right)=6\times \left(\frac{4-3}{12}\right)=\frac{1}{2}$$

Answered by CElcedricjunior last updated on 18/Jul/22

$$-\frac{1}{2}$$$$.....lecelebrecedricjunior......$$

Answered by Mathspace last updated on 18/Jul/22

$$f\left(x\right)=\frac{3}{1-\sqrt{x}}-\frac{2}{1{-}^3\sqrt{x}}$$$$wedothecha7gementx=1-t$$$$(sot\to 0)\Rightarrow$$$$f\left(x\right)=f(1-t)=\frac{3}{1-\sqrt{1-t}}-\frac{2}{1{-}^3\sqrt{x}}$$$${(1+u)}^{\alpha }\sim 1+\alpha u+\frac{\alpha (\alpha -1)}{2}{u}^2$$$${(1-u)}^{\alpha }\sim 1-\alpha u+\frac{\alpha (\alpha -1)}{2}{u}^2$$$${(1-t)}^{\frac{1}{2}}\sim 1-\frac{t}{2}-\frac{1}{8}{t}^2$$$${(1-t)}^{\frac{1}{3}}\sim 1-\frac{t}{3}-\frac{2}{18}{t}^2$$$$\Rightarrow f(1-t)\sim \frac{3}{\frac{t}{2}+\frac{t^2}{8}}-\frac{2}{\frac{t}{3}+\frac{t^2}{9}}$$$$=\frac{24}{4t+t^2}-\frac{18}{3t+t^2}$$$$=\frac{24(3t+t^2)-18(4t+t^2)}{(4t+t^2)(3t+t^2)}$$$$=\frac{72t+24t^2-72t-18t^2}{t^2(4+t)(3+t)}$$$$=\frac{6}{(4+t)(3+t)}\to \frac{6}{12}=\frac{1}{2}(t\to 0)$$

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