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Question Number 97117 by bobhans last updated on 06/Jun/20

$$\underset{\mathrm{h}\to 0}{\lim }\left[\frac{1}{\mathrm{h}}\underset{2}{\stackrel{2+2\mathrm{h}}{\int }}\sqrt{{\mathrm{t}}^2+2}\mathrm{dt}\right]$$

Answered by abdomathmax last updated on 06/Jun/20

$$\mathrm{let}\mathrm{use}\mathrm{hospital}\mathrm{theorem}\Rightarrow$$$${\lim }_{\mathrm{h}\to 0}\frac{{\int }_2^{2+2\mathrm{h}}\sqrt{{\mathrm{t}}^2+2}\mathrm{dt}}{\mathrm{h}}$$$$={\lim }_{\mathrm{h}\to 0}2\sqrt{{(2+2\mathrm{h})}^2+2}=2\sqrt{4+2}=2\sqrt{6}$$$$\mathrm{another}\mathrm{way}\mathrm{\exists }\mathrm{c}\in ]2,2+2\mathrm{h}[/\mathrm{A}\left(\mathrm{h}\right)=$$$$\frac{1}{\mathrm{h}}{\int }_2^{2+2\mathrm{h}}\sqrt{{\mathrm{t}}^2+2}\mathrm{dt}=\frac{1}{\mathrm{h}}\sqrt{{\mathrm{c}}^2+2}{\int }_2^{2+2\mathrm{h}}\mathrm{dt}$$$$=2\sqrt{{\mathrm{c}}^2+2}(\mathrm{h}\to 0)\Rightarrow \mathrm{c}\to 2\Rightarrow$$$${\lim }_{\mathrm{h}\to 0}\mathrm{A}\left(\mathrm{h}\right)=2\sqrt{6}$$

Commented by bobhans last updated on 07/Jun/20

$$\mathrm{yes}\mathrm{sir}.\mathrm{thank}\mathrm{you}$$

Commented by mathmax by abdo last updated on 07/Jun/20

$$\mathrm{you}\mathrm{are}\mathrm{welcome}$$

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