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Question Number 117551 by bobhans last updated on 12/Oct/20

$$\left(\mathrm{a}\right)\underset{x\to 1}{\lim }(\frac{1}{2(1-\sqrt{\mathrm{x}})}-\frac{1}{3(1-\sqrt[3]{\mathrm{x}})})=?$$$$\left(\mathrm{b}\right)\underset{x\to \mathrm{\infty }}{\lim }\frac{\ln (\mathrm{x}+\sqrt{1+{\mathrm{x}}^2})-\ln (\mathrm{x}+\sqrt{{\mathrm{x}}^2-1})}{{\left(\ln \left(\frac{\mathrm{x}+1}{\mathrm{x}-1}\right)\right)}^2}=?$$

Answered by bemath last updated on 13/Oct/20

$$\left(\mathrm{a}\right)\underset{x\to 1}{\lim }(\frac{1}{2(1-\sqrt{\mathrm{x}})}-\frac{1}{3(1-\sqrt[3]{\mathrm{x}})})=?$$$$\mathrm{letting}\mathrm{x}={\mathrm{r}}^6\to \{\begin{array}{l}\sqrt{\mathrm{x}}={\mathrm{r}}^3\\ \sqrt[3]{\mathrm{x}}={\mathrm{r}}^2\end{array}$$$$\underset{\mathrm{r}\to 1}{\lim }(\frac{1}{2(1-{\mathrm{r}}^3)}-\frac{1}{3(1-{\mathrm{r}}^2)})=$$$$\underset{\mathrm{r}\to 1}{\lim }(\frac{1}{2(1-\mathrm{r})(1+\mathrm{r}+{\mathrm{r}}^2)}-\frac{1}{3(1-\mathrm{r})(1+\mathrm{r})})=$$$$\underset{\mathrm{r}\to 1}{\lim }\left(\frac{3(1+\mathrm{r})-2(1+\mathrm{r}+{\mathrm{r}}^2)}{6(1-\mathrm{r})(1+\mathrm{r})(1+\mathrm{r}+{\mathrm{r}}^2)}\right)=$$$$\underset{\mathrm{r}\to 1}{\lim }\left(\frac{1+\mathrm{r}-{2\mathrm{r}}^2}{(1-\mathrm{r})}\right)\times \underset{\mathrm{r}\to 1}{\lim }\left(\frac{1}{6(1+\mathrm{r})(1+\mathrm{r}+{\mathrm{r}}^2)}\right)=$$$$\frac{1}{36}\times \underset{\mathrm{r}\to 1}{\lim }\left(\frac{1-4\mathrm{r}}{-1}\right)=\frac{1}{36}\times \left(3\right)=\frac{1}{12}$$

Answered by TANMAY PANACEA last updated on 12/Oct/20

$$a)t^6=x$$$$\underset{t\to 1}{\lim }(\frac{1}{2(1-t^3)}-\frac{1}{3(1-t^2)})$$$$\underset{t\to 1}{\lim }[\frac{1}{2(1-t)(1+t+t^2)}-\frac{1}{3(1+t)(1-t)}]$$$$=\underset{t\to 1}{\lim }\left[\frac{3(1+t)-2(1+t+t^2)}{6(1-t)(1+t+t^2)(1+t)}\right]$$$$=\underset{t\to 1}{\lim }\left[\frac{3+3t-2-2t-2t^2}{6(1-t)(1+t+t^2)(1+t)}\right]$$$$=\underset{t\to 1}{\lim }\left[\frac{1+t-2t^2}{6(1-t)(1+t+t^2)(1+t)}\right]$$$$=\underset{t\to 1}{\lim }\left[\frac{(1+2t)(1-t)}{6(1-t)(1+t+t^2)(1+t)}\right]=\frac{3}{6\times 3\times 2}=\frac{1}{12}$$

Commented by TANMAY PANACEA last updated on 12/Oct/20

$$plscheckmistakeifany$$

Answered by TANMAY PANACEA last updated on 12/Oct/20

$$\underset{x\to \mathrm{\infty }}{\lim }\frac{ln\left(\frac{x+\sqrt{1+x^2}}{x+\sqrt{x^2-1}}\right)}{{\left(ln\left(\frac{x+1}{x-1}\right)\right)}^2}=\underset{x\to \mathrm{\infty }}{\lim }\frac{ln\left(\frac{1+\sqrt{\frac{1}{x^2}+1}}{1+\sqrt{1-\frac{1}{x^2}}}\right)}{{\left\{ln\left(\frac{1+\frac{1}{x}}{1-\frac{1}{x}}\right)\right\}}^2}=\underset{x\to \mathrm{\infty }}{\mathrm{li}}$$$$=\frac{ln1}{ln1}=\frac{0}{0}$$

Commented by MJS_new last updated on 12/Oct/20

$$\mathrm{approximating}\mathrm{I}\mathrm{get}\frac{1}{8}$$

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