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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

$$1)case1{lim}_{x\to 0^{+}}\frac{x}{\mid x\mid +x^2}={lim}_{x\to 0^{+}}\frac{x}{x+x^2}={lim}_{x\to 0^{+}}\frac{1}{1+x}=1$$$$case2{lim}_{x\to \stackrel{-}{0}}\frac{x}{\mid x\mid +x^2}={lim}_{x\to 0^{-}}\frac{x}{-x+x^2}={lim}_{x\to 0^{-}}\frac{1}{x-1}=-1so$$$$$$

Commented by tanmay.chaudhury50@gmail.com last updated on 29/Dec/18

$$thankyousir...$$

Answered by tanmay.chaudhury50@gmail.com last updated on 30/Dec/18

$$ii)sincex<0\mid x\mid =-x$$$$\underset{x\to -\mathrm{\infty }}{\lim }\frac{x^{5}tan\left(\frac{1}{\pi x^2}\right)+3\mid x{\mid }^2+7}{\mid x{\mid }^3+7\mid x\mid +8}$$$$=\underset{x\to -\mathrm{\infty }}{\lim }\frac{x^{5}tan\left(\frac{1}{\pi x^2}\right)+3x^2+7}{-x^3-7x+8}$$$$=\underset{x\to -\mathrm{\infty }}{\lim }\frac{x^{2}tan\left(\frac{1}{\pi x^2}\right)+\frac{3}{x}+\frac{7}{x^3}}{-1-\frac{7}{x^2}+\frac{8}{x^3}}[x=\frac{1}{k}]$$$$=\underset{k\to 0}{\lim }\frac{\frac{1}{\pi }\times \left[\frac{sin\left(\frac{k^2}{\pi }\right)}{\frac{k^2}{\pi }}\right]\times \frac{1}{cos\left(\frac{k^2}{\pi }\right)}+3k+7k^3}{-1-7k^2+8k^3}$$$$=\frac{\frac{1}{\pi }\times 1+0+0}{-1-0-0}=\frac{-1}{\pi }$$

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