Question Number 136278 by adhigenz last updated on 20/Mar/21
![How to solve this? lim_(x→∞) ((∫_1 ^x [(√(x^2 −4x)) − x] dx)/x^2 ) = ...](https://www.tinkutara.com/question/Q136278.png)
$$\mathrm{How}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{this}? \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\underset{\mathrm{1}} {\overset{{x}} {\int}}\left[\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}}\:−\:{x}\right]\:{dx}}{{x}^{\mathrm{2}} }\:=\:… \\ $$
Answered by MJS_new last updated on 20/Mar/21
$$\mathrm{error}:\:\mathrm{an}\:\mathrm{integral}\:\mathrm{cannot}\:\mathrm{have}\:\mathrm{dependent} \\ $$$$\mathrm{borders}.\:\mathrm{we}\:\mathrm{cannot}\:\mathrm{solve}\:\underset{{a}} {\overset{{x}} {\int}}{f}\left({x}\right){dx} \\ $$
Commented by ajfour last updated on 20/Mar/21
$${no}\:{error},\:{it}\:{should}\:{be}\:{same}\:{as} \\ $$$$\int_{{a}} ^{\:{x}} {f}\left({t}\right){dt}\:\:. \\ $$
Answered by mathmax by abdo last updated on 20/Mar/21
$$\mathrm{i}\:\mathrm{think}\:\mathrm{the}\:\mathrm{Q}\:\mathrm{is}\:\mathrm{lim}_{\mathrm{x}\rightarrow\infty} \:\:\frac{\int_{\mathrm{1}} ^{\mathrm{x}} \left(\sqrt{\mathrm{t}^{\mathrm{2}} −\mathrm{4t}}−\mathrm{t}\right)\mathrm{dt}}{\mathrm{x}^{\mathrm{2}} } \\ $$
Commented by mathmax by abdo last updated on 20/Mar/21
$$\mathrm{L}=_{\mathrm{hospital}} \:\:\:\:\mathrm{lim}_{\mathrm{x}\rightarrow\infty} \:\:\:\:\frac{\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{4x}}−\mathrm{x}}{\mathrm{2x}} \\ $$$$=\mathrm{lim}_{\mathrm{x}\rightarrow\infty} \:\:\:\:\frac{\left(\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{4x}}−\mathrm{x}\right)\left(\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{4x}}+\mathrm{x}\right)}{\mathrm{2x}\left(\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{4x}}+\mathrm{x}\right)} \\ $$$$=\mathrm{lim}_{\mathrm{x}\rightarrow\infty} \:\:\:\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{4x}−\mathrm{x}^{\mathrm{2}} }{\mathrm{2x}\left(\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{4x}}+\mathrm{x}\right)}\:=\mathrm{lim}_{\mathrm{x}\rightarrow\infty} \:\:\:\:\frac{−\mathrm{2}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{4x}}+\mathrm{x}}=\mathrm{0} \\ $$