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Question Number 136278 by adhigenz last updated on 20/Mar/21

$$\mathrm{How}\mathrm{to}\mathrm{solve}\mathrm{this}?$$$$\underset{x\to \mathrm{\infty }}{\lim }\frac{\underset{1}{\stackrel{x}{\int }}[\sqrt{x^2-4x}-x]dx}{x^2}=...$$

Answered by MJS_new last updated on 20/Mar/21

$$\mathrm{error}:\mathrm{an}\mathrm{integral}\mathrm{cannot}\mathrm{have}\mathrm{dependent}$$$$\mathrm{borders}.\mathrm{we}\mathrm{cannot}\mathrm{solve}\underset{a}{\stackrel{x}{\int }}f\left(x\right)dx$$

Commented by ajfour last updated on 20/Mar/21

$$noerror,itshouldbesameas$$$${\int }_a^{x}f\left(t\right)dt.$$

Answered by mathmax by abdo last updated on 20/Mar/21

$$\mathrm{i}\mathrm{think}\mathrm{the}\mathrm{Q}\mathrm{is}{\lim }_{\mathrm{x}\to \mathrm{\infty }}\frac{{\int }_1^{\mathrm{x}}(\sqrt{{\mathrm{t}}^2-4\mathrm{t}}-\mathrm{t})\mathrm{dt}}{{\mathrm{x}}^2}$$

Commented by mathmax by abdo last updated on 20/Mar/21

$$\mathrm{L}{=}_{\mathrm{hospital}}{\lim }_{\mathrm{x}\to \mathrm{\infty }}\frac{\sqrt{{\mathrm{x}}^2-4\mathrm{x}}-\mathrm{x}}{2\mathrm{x}}$$$$={\lim }_{\mathrm{x}\to \mathrm{\infty }}\frac{(\sqrt{{\mathrm{x}}^2-4\mathrm{x}}-\mathrm{x})(\sqrt{{\mathrm{x}}^2-4\mathrm{x}}+\mathrm{x})}{2\mathrm{x}(\sqrt{{\mathrm{x}}^2-4\mathrm{x}}+\mathrm{x})}$$$$={\lim }_{\mathrm{x}\to \mathrm{\infty }}\frac{{\mathrm{x}}^2-4\mathrm{x}-{\mathrm{x}}^2}{2\mathrm{x}(\sqrt{{\mathrm{x}}^2-4\mathrm{x}}+\mathrm{x})}={\lim }_{\mathrm{x}\to \mathrm{\infty }}\frac{-2}{\sqrt{{\mathrm{x}}^2-4\mathrm{x}}+\mathrm{x}}=0$$

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