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Question Number 46330 by Saorey last updated on 24/Oct/18

$$\mathrm{pls}\mathrm{help}\mathrm{me}!$$$$\mathrm{L}=\underset{x\to 1}{\lim }\frac{\sqrt[13]{\mathrm{x}}-\sqrt[7]{\mathrm{x}}}{\sqrt[5]{\mathrm{x}}-\sqrt[3]{\mathrm{x}}}$$

Commented by MJS last updated on 24/Oct/18

$$\mathrm{I}\mathrm{would}\mathrm{again}\mathrm{recommend}{\mathrm{l}}^{\prime }\mathrm{Hopital}$$$$\underset{x\to 1}{\lim }\frac{x^a-x^b}{x^c-x^d}=\underset{x\to 1}{\lim }\frac{\frac{d}{dx}[x^a-x^b]}{\frac{d}{dx}[x^c-x^d]}=\underset{x\to 1}{\lim }\frac{{ax}^{a-1}-{bx}^{b-1}}{{cx}^{c-1}-{dx}^{d-1}}=$$$$=\underset{x\to 1}{\lim }\frac{{ax}^a-{bx}^b}{{cx}^c-{dx}^d}=\frac{a-b}{c-d}$$$$\mathrm{with}a=\frac{1}{13};b=\frac{1}{7};c=\frac{1}{5};d=\frac{1}{3}$$$$L=\frac{a-b}{c-d}=\frac{\frac{1}{13}-\frac{1}{7}}{\frac{1}{5}-\frac{1}{3}}=\frac{-\frac{6}{91}}{-\frac{2}{15}}=\frac{45}{91}$$

Commented by hassentimol last updated on 24/Oct/18

$$Whatis{L}^{\prime }Hopital,pleasesir?$$$$$$

Commented by MJS last updated on 24/Oct/18

$$\underset{x\to a}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to a}{\lim }\frac{f^{\prime }\left(x\right)}{g^{\prime }\left(x\right)}\mathrm{if}f\left(a\right)=0\mathrm{and}g\left(a\right)=0\mathrm{xor}$$$$f\left(a\right)=\pm \mathrm{\infty }\mathrm{and}g\left(a\right)=\pm \mathrm{\infty }$$$$\mathrm{example}$$$$\underset{x\to 0}{\lim }\left(x\ln x\right)=\underset{x\to 0}{\lim }\frac{\ln x}{x^{-1}}=\underset{x\to 0}{\lim }\frac{\frac{d}{dx}\left[\ln x\right]}{\frac{d}{dx}\left[x^{-1}\right]}=$$$$=\underset{x\to 0}{\lim }\frac{x^{-1}}{-x^{-2}}=\underset{x\to 0}{\lim }-x=0$$

Answered by ajfour last updated on 24/Oct/18

$$L=\underset{h\to 0}{\lim }\frac{{(1+h)}^{1/13}-{(1+h)}^{1/7}}{{(1+h)}^{1/5}-{(1+h)}^{1/3}}$$$$=\underset{h\to 0}{\lim }\frac{1+\frac{h}{13}+...-1-\frac{h}{7}-...}{1+\frac{h}{5}+...-1-\frac{h}{3}-...}$$$$=\frac{-6/91}{-2/15}=\frac{45}{91}.$$

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