Menu Close

Question-216491

Tinku Tara 0 Comments
Pin




Question Number 216491 by Jubr last updated on 09/Feb/25
Answered by MrGaster last updated on 09/Feb/25
(1): =lim_(x→0) ((x∫_0 ^x (1−t^2 +(t^4 /(2!))−(t^6 /(3!))+…)dt)/( (√(1−e^(−x^2 ) )))) =lim_(x→0) ((x(x−(x^3 /3)+(x^5 /(10))−(x^7 /(3!))+…)dt)/( (√(1−e^(−x^2 ) )))) =lim_(x→0) ((x^2 −(x^4 /3)+(x^6 /(10))−(x^8 /(42))+…)/( (√(1−e^(−x^2 ) )))) =lim_(x→0) (x^2 /( (√(1−(1−(x^2 /2)+(x^4 /8)−(x^6 /(48))+…))))) =lim_(x→0) (x^2 /( (√(1−(x^2 /2)−(x^4 /8)+(x^6 /(24))−…)))) =lim_(x→0) ((x(√2))/( (√(1−(x^2 /4)+(x^4 /(24))−…)))) =lim_(x→0) x(√2) = determinant ((0)) (2): ∫_0 ^x e^(−t^2 ) dt≈x−(x^3 /3)+(x^5 /(10))−… x∫_0 ^x e^(−t^2 ) dt≈x^2 −(x^4 /3)+(x^6 /(10))−… (√(1−e^(−x^2 ) ))≈(√(x^2 −(x^4 /2)+…))≈x(1−(x^2 /4)+…) ((x∫_0 ^x e^(−t^2 ) dt)/( (√(1−e^(−x^2 ) ))))≈((x^2 −(x^4 /3))/(x(1−(x^2 /4))))≈((x(1−(x^2 /3)))/(1−(x^2 /4))) ≈x(1−(x^2 /(12)))→ determinant ((0))
$$\left(\mathrm{1}\right): \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}\int_{\mathrm{0}} ^{{x}} \left(\mathrm{1}−{t}^{\mathrm{2}} +\frac{{t}^{\mathrm{4}} }{\mathrm{2}!}−\frac{{t}^{\mathrm{6}} }{\mathrm{3}!}+\ldots\right){dt}}{\:\sqrt{\mathrm{1}−{e}^{−{x}^{\mathrm{2}} } }} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}\left({x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{{x}^{\mathrm{5}} }{\mathrm{10}}−\frac{{x}^{\mathrm{7}} }{\mathrm{3}!}+\ldots\right){dt}}{\:\sqrt{\mathrm{1}−{e}^{−{x}^{\mathrm{2}} } }} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}^{\mathrm{2}} −\frac{{x}^{\mathrm{4}} }{\mathrm{3}}+\frac{{x}^{\mathrm{6}} }{\mathrm{10}}−\frac{{x}^{\mathrm{8}} }{\mathrm{42}}+\ldots}{\:\sqrt{\mathrm{1}−{e}^{−{x}^{\mathrm{2}} } }} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{4}} }{\mathrm{8}}−\frac{{x}^{\mathrm{6}} }{\mathrm{48}}+\ldots\right)}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\frac{{x}^{\mathrm{4}} }{\mathrm{8}}+\frac{{x}^{\mathrm{6}} }{\mathrm{24}}−\ldots}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{4}}+\frac{{x}^{\mathrm{4}} }{\mathrm{24}}−\ldots}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{x}\sqrt{\mathrm{2}} \\ $$$$=\begin{array}{|c|}{\mathrm{0}}\\\hline\end{array} \\ $$$$\left(\mathrm{2}\right): \\ $$$$\int_{\mathrm{0}} ^{{x}} {e}^{−{t}^{\mathrm{2}} } {dt}\approx{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{{x}^{\mathrm{5}} }{\mathrm{10}}−\ldots \\ $$$${x}\int_{\mathrm{0}} ^{{x}} {e}^{−{t}^{\mathrm{2}} } {dt}\approx{x}^{\mathrm{2}} −\frac{{x}^{\mathrm{4}} }{\mathrm{3}}+\frac{{x}^{\mathrm{6}} }{\mathrm{10}}−\ldots \\ $$$$\sqrt{\mathrm{1}−{e}^{−{x}^{\mathrm{2}} } }\approx\sqrt{{x}^{\mathrm{2}} −\frac{{x}^{\mathrm{4}} }{\mathrm{2}}+\ldots}\approx{x}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{4}}+\ldots\right) \\ $$$$\frac{{x}\int_{\mathrm{0}} ^{{x}} {e}^{−{t}^{\mathrm{2}} } {dt}}{\:\sqrt{\mathrm{1}−{e}^{−{x}^{\mathrm{2}} } }}\approx\frac{{x}^{\mathrm{2}} −\frac{{x}^{\mathrm{4}} }{\mathrm{3}}}{{x}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{4}}\right)}\approx\frac{{x}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{3}}\right)}{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{4}}} \\ $$$$\approx{x}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{12}}\right)\rightarrow\begin{array}{|c|}{\mathrm{0}}\\\hline\end{array} \\ $$
Commented by Jubr last updated on 09/Feb/25
Thanks sir.
$${Thanks}\:{sir}. \\ $$
Answered by mathmax last updated on 10/Feb/25
u(x)=x∫_0 ^x e^(−t^2 ) dt et v(x)=(√(1−e^(−x^2 ) )) lim_(x→0) u(x)=lim_(x→0) v(x)=0 on applique l hospital l=lim_(x→0) ((u^′ (x))/(v^′ (x)))=lim_(x→0) ((∫_0 ^x e^(−t^2 ) dt+ xe^(−x^2 ) )/((2xe^(−x^2 ) )/(2(√(1−e^(−x^2 ) ))))) =lim_(x→0) (((√(1−e^(−x^2 ) ))(∫_0 ^x e^(−t^2 ) dt+xe^(−x^2 ) ))/(xe^(−x^2 ) )) we have e^(−x^2 ) ∼1−x^2 ⇒1−e^(−x^2 ) ∼x^2 and (√(1−e^(−x^2 ) ))∼x ⇒ l=lim_(x→0) ((x(∫_0 ^x e^(−t^2 ) dt+xe^(−x^2 ) ))/(xe^(−x^2 ) )) =lim_(x→0) ((∫_0 ^x e^(−t^2 ) dt+xe^(−x^2 ) )/e^(−x^2 ) )=(o/1)=0
$${u}\left({x}\right)={x}\int_{\mathrm{0}} ^{{x}} \:{e}^{−{t}^{\mathrm{2}} } {dt}\:{et}\:{v}\left({x}\right)=\sqrt{\mathrm{1}−{e}^{−{x}^{\mathrm{2}} } } \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} {u}\left({x}\right)={lim}_{{x}\rightarrow\mathrm{0}} {v}\left({x}\right)=\mathrm{0}\:{on}\:{applique}\:{l}\:{hospital} \\ $$$${l}={lim}_{{x}\rightarrow\mathrm{0}} \:\:\frac{{u}^{'} \left({x}\right)}{{v}^{'} \left({x}\right)}={lim}_{{x}\rightarrow\mathrm{0}} \frac{\int_{\mathrm{0}} ^{{x}} {e}^{−{t}^{\mathrm{2}} } {dt}+\:{xe}^{−{x}^{\mathrm{2}} } }{\frac{\mathrm{2}{xe}^{−{x}^{\mathrm{2}} } }{\mathrm{2}\sqrt{\mathrm{1}−{e}^{−{x}^{\mathrm{2}} } }}} \\ $$$$={lim}_{{x}\rightarrow\mathrm{0}} \frac{\sqrt{\mathrm{1}−{e}^{−{x}^{\mathrm{2}} } }\left(\int_{\mathrm{0}} ^{{x}} \:{e}^{−{t}^{\mathrm{2}} } {dt}+{xe}^{−{x}^{\mathrm{2}} } \right)}{{xe}^{−{x}^{\mathrm{2}} } } \\ $$$${we}\:{have}\:\:{e}^{−{x}^{\mathrm{2}} } \sim\mathrm{1}−{x}^{\mathrm{2}} \:\Rightarrow\mathrm{1}−{e}^{−{x}^{\mathrm{2}} } \sim{x}^{\mathrm{2}} \\ $$$${and}\:\sqrt{\mathrm{1}−{e}^{−{x}^{\mathrm{2}} } }\sim{x}\:\Rightarrow \\ $$$${l}={lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\frac{{x}\left(\int_{\mathrm{0}} ^{{x}} {e}^{−{t}^{\mathrm{2}} } {dt}+{xe}^{−{x}^{\mathrm{2}} } \right)}{{xe}^{−{x}^{\mathrm{2}} } } \\ $$$$={lim}_{{x}\rightarrow\mathrm{0}} \:\:\frac{\int_{\mathrm{0}} ^{{x}} {e}^{−{t}^{\mathrm{2}} } {dt}+{xe}^{−{x}^{\mathrm{2}} } }{{e}^{−{x}^{\mathrm{2}} } }=\frac{{o}}{\mathrm{1}}=\mathrm{0} \\ $$
Commented by Jubr last updated on 10/Feb/25
Thanks sir.
$${Thanks}\:{sir}. \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *