Question Number 51630 by tanmay.chaudhury50@gmail.com last updated on 29/Dec/18
Commented by tanmay.chaudhury50@gmail.com last updated on 29/Dec/18
Commented by tanmay.chaudhury50@gmail.com last updated on 29/Dec/18
Commented by maxmathsup by imad last updated on 29/Dec/18
$$\left.\mathrm{1}\right){case}\:\mathrm{1}\:\:\:\:{lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:\:\:\:\frac{{x}}{\mid{x}\mid+{x}^{\mathrm{2}} }\:={lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:\:\:\frac{{x}}{{x}+{x}^{\mathrm{2}} }\:={lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+{x}}\:=\mathrm{1} \\ $$$${case}\:\mathrm{2}\:\:\:{lim}_{{x}\rightarrow\overset{−} {\mathrm{0}}} \:\:\:\:\:\frac{{x}}{\mid{x}\mid+{x}^{\mathrm{2}} }\:={lim}_{{x}\rightarrow\mathrm{0}^{−} } \:\:\frac{{x}}{−{x}+{x}^{\mathrm{2}} }\:={lim}_{{x}\rightarrow\mathrm{0}^{−} } \:\:\:\:\frac{\mathrm{1}}{{x}−\mathrm{1}}\:=−\mathrm{1}\:{so} \\ $$$$ \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 29/Dec/18
$${thank}\:{you}\:{sir}… \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 30/Dec/18
![ii)since x<0 ∣x∣=−x lim_(x→−∞) ((x^5 tan((1/(πx^2 )))+3∣x∣^2 +7)/(∣x∣^3 +7∣x∣+8)) =lim_(x→−∞) ((x^5 tan((1/(πx^2 )))+3x^2 +7)/(−x^3 −7x+8)) =lim_(x→−∞) ((x^2 tan((1/(πx^2 )))+(3/x)+(7/x^3 ))/(−1−(7/x^2 )+(8/x^3 ))) [x=(1/k)] =lim_(k→0) (((1/π)×[((sin((k^2 /π)))/(k^2 /π))]×(1/(cos((k^2 /π))))+3k+7k^3 )/(−1−7k^2 +8k^3 )) =(((1/π)×1+0+0)/(−1−0−0))=((−1)/π)](https://www.tinkutara.com/question/Q51793.png)
$$\left.{ii}\right){since}\:{x}<\mathrm{0}\:\:\:\:\:\mid{x}\mid=−{x} \\ $$$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\frac{{x}^{\mathrm{5}} {tan}\left(\frac{\mathrm{1}}{\pi{x}^{\mathrm{2}} }\right)+\mathrm{3}\mid{x}\mid^{\mathrm{2}} +\mathrm{7}}{\mid{x}\mid^{\mathrm{3}} +\mathrm{7}\mid{x}\mid+\mathrm{8}} \\ $$$$=\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\frac{{x}^{\mathrm{5}} {tan}\left(\frac{\mathrm{1}}{\pi{x}^{\mathrm{2}} }\right)+\mathrm{3}{x}^{\mathrm{2}} +\mathrm{7}}{−{x}^{\mathrm{3}} −\mathrm{7}{x}+\mathrm{8}} \\ $$$$=\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\frac{{x}^{\mathrm{2}} {tan}\left(\frac{\mathrm{1}}{\pi{x}^{\mathrm{2}} }\right)+\frac{\mathrm{3}}{{x}}+\frac{\mathrm{7}}{{x}^{\mathrm{3}} }}{−\mathrm{1}−\frac{\mathrm{7}}{{x}^{\mathrm{2}} }+\frac{\mathrm{8}}{{x}^{\mathrm{3}} }}\:\:\:\:\:\:\:\:\:\left[{x}=\frac{\mathrm{1}}{{k}}\right] \\ $$$$=\underset{{k}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{\mathrm{1}}{\pi}×\left[\frac{{sin}\left(\frac{{k}^{\mathrm{2}} }{\pi}\right)}{\frac{{k}^{\mathrm{2}} }{\pi}}\right]×\frac{\mathrm{1}}{{cos}\left(\frac{{k}^{\mathrm{2}} }{\pi}\right)}+\mathrm{3}{k}+\mathrm{7}{k}^{\mathrm{3}} }{−\mathrm{1}−\mathrm{7}{k}^{\mathrm{2}} +\mathrm{8}{k}^{\mathrm{3}} } \\ $$$$=\frac{\frac{\mathrm{1}}{\pi}×\mathrm{1}+\mathrm{0}+\mathrm{0}}{−\mathrm{1}−\mathrm{0}−\mathrm{0}}=\frac{−\mathrm{1}}{\pi} \\ $$