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Question Number 104533 by 175mohamed last updated on 22/Jul/20

$$find:$$$$\underset{x\to 1}{\lim }\frac{{(x+2)}^2+{(x+1)}^3-17}{\sqrt{x+3}-\sqrt[3]{x+7}}$$

Answered by Dwaipayan Shikari last updated on 22/Jul/20

$$\underset{x\to 1}{\lim }\frac{2(x+2)+3{(x+1)}^2}{\frac{1}{2}.\frac{1}{\sqrt{x+3}}-\frac{1}{3}{(x+7)}^{-\frac{2}{3}}}=\frac{18}{\frac{1}{4}-\frac{1}{12}}=108$$

Answered by OlafThorendsen last updated on 22/Jul/20

$$x=u+1$$$$\underset{u\to 0}{\lim }\frac{{(u+3)}^2+{(u+2)}^3-17}{\sqrt{u+4}-\sqrt[3]{u+8}}$$$$\underset{u\to 0}{\lim }\frac{9{(1+\frac{u}{3})}^2+8{(1+\frac{u}{2})}^3-17}{2\sqrt{1+\frac{u}{4}}-2\sqrt[3]{1+\frac{u}{8}}}$$$$\underset{u\to 0}{\lim }\frac{9(1+\frac{2u}{3})+8(1+\frac{3u}{2})-17}{2(1+\frac{u}{8})-2(1+\frac{u}{24})}$$$$\underset{u\to 0}{\lim }\frac{9\left(\frac{2u}{3}\right)+8\left(\frac{3u}{2}\right)}{2\left(\frac{u}{8}\right)-2\left(\frac{u}{24}\right)}$$$$\underset{u\to 0}{\lim }\frac{6u+12u}{\frac{u}{4}-\frac{u}{12}}$$$$\underset{u\to 0}{\lim }\frac{18u}{\frac{u}{6}}=18\times 6=108$$$$$$

Commented by bobhans last updated on 22/Jul/20

$$\frac{u}{4}-\frac{u}{12}=\frac{3u}{12}-\frac{u}{12}=\frac{2u}{12}=\frac{u}{6}$$

Answered by bramlex last updated on 22/Jul/20

$$L^{\prime }Hopitalrule$$$$\underset{x\to 1}{\lim }\frac{2(x+2)+3{(x+1)}^2}{\frac{1}{2\sqrt{x+3}}-\frac{1}{3\sqrt[3]{{(x+7)}^2}}}=$$$$\frac{6+12}{\frac{1}{4}-\frac{1}{3.4}}=\frac{18}{\left(\frac{1}{6}\right)}=108◻$$

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