Question-113341

Question Number 113341 by Khalmohmmad last updated on 12/Sep/20

Answered by mathmax by abdo last updated on 13/Sep/20

we have ∣sin((π/x))∣≤1 ⇒∣(√(x^3 +x^2 ))sin((π/x))∣≤∣x∣(√(x+1)) →0 (x→0) ⇒lim_(x→0) (√(x^3 +x^2 ))sin((π/x))=0

$$\mathrm{we}\:\mathrm{have}\:\mid\mathrm{sin}\left(\frac{\pi}{\mathrm{x}}\right)\mid\leqslant\mathrm{1}\:\Rightarrow\mid\sqrt{\mathrm{x}^{\mathrm{3}} \:+\mathrm{x}^{\mathrm{2}} }\mathrm{sin}\left(\frac{\pi}{\mathrm{x}}\right)\mid\leqslant\mid\mathrm{x}\mid\sqrt{\mathrm{x}+\mathrm{1}}\:\rightarrow\mathrm{0}\:\left(\mathrm{x}\rightarrow\mathrm{0}\right) \\ $$$$\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\:\:\:\:\sqrt{\mathrm{x}^{\mathrm{3}} \:+\mathrm{x}^{\mathrm{2}} }\mathrm{sin}\left(\frac{\pi}{\mathrm{x}}\right)=\mathrm{0} \\ $$

Answered by RCRC last updated on 12/Sep/20

lim_(x→0) (√(x^3 +x^2 )) sen((π/x)) = lim_(x→0) ( ((πx (√(x+1)))/π) sen((π/x))) = lim_(x→0) ( ((πx (√(x+1)))/π) sen((π/x))) = lim_(x→0) ( ((π (√(x+1)))/(πx^(−1) )) sen(πx^(−1) )) =lim_(x→0) (π (√(x+1)) ((sen(πx^(−1) ))/(πx^(−1) ))) =lim_(x→0) (π (√(x+1)))×lim_(x→0) (((sen(πx^(−1) ))/(πx^(−1) )))∡ =π×0=0

$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\sqrt{{x}^{\mathrm{3}} +{x}^{\mathrm{2}} }\:{sen}\left(\frac{\pi}{{x}}\right)\:=\:\underset{{x}\rightarrow\mathrm{0}} {{lim}}\left(\:\frac{\pi{x}\:\sqrt{{x}+\mathrm{1}}}{\pi}\:{sen}\left(\frac{\pi}{{x}}\right)\right) \\ $$$$\:=\:\underset{{x}\rightarrow\mathrm{0}} {{lim}}\left(\:\frac{\pi{x}\:\sqrt{{x}+\mathrm{1}}}{\pi}\:{sen}\left(\frac{\pi}{{x}}\right)\right) \\ $$$$\:=\:\underset{{x}\rightarrow\mathrm{0}} {{lim}}\left(\:\frac{\pi\:\sqrt{{x}+\mathrm{1}}}{\pi{x}^{−\mathrm{1}} }\:{sen}\left(\pi{x}^{−\mathrm{1}} \right)\right) \\ $$$$\:=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\left(\pi\:\sqrt{{x}+\mathrm{1}}\:\frac{{sen}\left(\pi{x}^{−\mathrm{1}} \right)}{\pi{x}^{−\mathrm{1}} }\right) \\ $$$$\:=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\left(\pi\:\sqrt{{x}+\mathrm{1}}\right)×\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\left(\frac{{sen}\left(\pi{x}^{−\mathrm{1}} \right)}{\pi{x}^{−\mathrm{1}} }\right)\measuredangle \\ $$$$\:=\pi×\mathrm{0}=\mathrm{0} \\ $$

Answered by Dwaipayan Shikari last updated on 12/Sep/20

lim_(x→0) (√(x^3 +x)) sin((π/x))=lim_(x→0) (√(x^3 +x)) sin(z) (z→∞) −1≤sin(z)≤1 so (√(x^3 +x)) sin(z)=0

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\sqrt{{x}^{\mathrm{3}} +{x}}\:\:{sin}\left(\frac{\pi}{{x}}\right)=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\sqrt{{x}^{\mathrm{3}} +{x}}\:\:{sin}\left({z}\right)\:\:\left({z}\rightarrow\infty\right) \\ $$$$−\mathrm{1}\leqslant{sin}\left({z}\right)\leqslant\mathrm{1} \\ $$$${so}\:\:\sqrt{{x}^{\mathrm{3}} +{x}}\:\:{sin}\left({z}\right)=\mathrm{0} \\ $$

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