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Calculate-lim-x-0-2x-1-2x-1-e-t-2-dt-1-1-e-t-2-dt-x-2-8e-Don-t-use-L-Hospital-s-rule-




Question Number 168696 by qaz last updated on 16/Apr/22
Calculate   ::  lim_(x→0) ((∫_(2x−1) ^(2x+1) e^t^2  dt−∫_(−1) ^1 e^t^2  dt)/x^2 )=8e   ,Don′t use L′Hospital′s rule.
$$\mathrm{Calculate}\:\:\:::\:\:\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\int_{\mathrm{2x}−\mathrm{1}} ^{\mathrm{2x}+\mathrm{1}} \mathrm{e}^{\mathrm{t}^{\mathrm{2}} } \mathrm{dt}−\int_{−\mathrm{1}} ^{\mathrm{1}} \mathrm{e}^{\mathrm{t}^{\mathrm{2}} } \mathrm{dt}}{\mathrm{x}^{\mathrm{2}} }=\mathrm{8e}\:\:\:,\mathrm{Don}'\mathrm{t}\:\mathrm{use}\:\mathrm{L}'\mathrm{Hospital}'\mathrm{s}\:\mathrm{rule}. \\ $$
Answered by aleks041103 last updated on 17/Apr/22
let f(x)=∫_(2x−1) ^( 2x+1) e^t^2  dt  ⇒f(x)=f(0)+f′(0)x+(1/2)f′′(0)x^2 +...
$${let}\:{f}\left({x}\right)=\int_{\mathrm{2}{x}−\mathrm{1}} ^{\:\mathrm{2}{x}+\mathrm{1}} {e}^{{t}^{\mathrm{2}} } {dt} \\ $$$$\Rightarrow{f}\left({x}\right)={f}\left(\mathrm{0}\right)+{f}'\left(\mathrm{0}\right){x}+\frac{\mathrm{1}}{\mathrm{2}}{f}''\left(\mathrm{0}\right){x}^{\mathrm{2}} +… \\ $$

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