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Question Number 28770 by Rasheed.Sindhi last updated on 30/Jan/18

$$\underset{x\to +\mathrm{\infty }}{lim}\frac{5x^4-10x^2+1}{-3x^3+10x^2+50}$$

Commented by Cheyboy last updated on 30/Jan/18

$$\underset{x\to +\mathrm{\infty }}{\lim }\frac{\frac{5x^4}{x^4}-\frac{10x^2}{x^4}-\frac{1}{x^4}}{\frac{-3x^3}{x^4}-\frac{10x^2}{x^4}-\frac{50}{x^4}}$$$$\underset{x\to +\mathrm{\infty }}{\lim }\frac{5-0-0}{0-0-0}=5$$$$ievenforgetthat\frac{5}{0}=\mathrm{\infty }$$$$somyanswerseemscorrect$$

Commented by abdo imad last updated on 30/Jan/18

$$youranserisnotcorrect?cheyboygenalyif$$$$p\left(x\right)=a_n{x}^n+a_{n-1}{x}^{n-1}+....a_{0}and$$$$q\left(x\right)=b_m{x}^m+b_{m-1}{x}^{m-1}+...+b_{o}with{a}_n\ne 0and{b}_m\ne 0$$$${lim}_{x\to +\mathrm{\infty }}\frac{p\left(x\right)}{q\left(x\right)}={lim}_{x\to +\mathrm{\infty }}\frac{a_n{x}^n}{b_m{x}^m}and$$$${lim}_{x\to -\mathrm{\infty }}\frac{p\left(x\right)}{q\left(x\right)}={lim}_{x\to -\mathrm{\infty }}\frac{a_n{x}^n}{b_m{x}^m}forthat$$$${lim}_{x\to +\mathrm{\infty }}\frac{5x^4-10x^2-1}{-3x^3-10x^2-50}={lim}_{x\to +\mathrm{\infty }}\frac{5x^4}{-3x^3}$$$$={lim}_{x\to +\mathrm{\infty }}-\frac{5}{3}x=-\mathrm{\infty }.$$

Commented by abdo imad last updated on 30/Jan/18

$$={lim}_{x\to +\mathrm{\infty }}\frac{5x^4}{-3x^3}={lim}_{x\to +\mathrm{\infty }}\frac{-5}{3}x=-\mathrm{\infty }.$$

Commented by Cheyboy last updated on 30/Jan/18

$$Thanksirfortherectification$$$$$$$$$$

Commented by Cheyboy last updated on 30/Jan/18

$$Thanksirfortherectification$$$$$$$$$$

Commented by Rasheed.Sindhi last updated on 30/Jan/18

$$\mathcal{T}hankyouboth!$$

Answered by ajfour last updated on 30/Jan/18

$$=\underset{x\to +\mathrm{\infty }}{\lim }x\left[\frac{5x^3-10x-\frac{1}{x}}{-3x^3+10x^2+50}\right]$$$$=\left(\underset{x\to +\mathrm{\infty }}{\lim }x\right)\times \underset{x\to +\mathrm{\infty }}{\lim }\left(\frac{5-\frac{10}{x^2}-\frac{1}{x^3}}{-3+\frac{10}{x}+\frac{50}{x^3}}\right)$$$$=-\frac{5}{3}\underset{x\to +\mathrm{\infty }}{\lim }\left(x\right)=-\mathrm{\infty }.$$

Commented by Rasheed.Sindhi last updated on 30/Jan/18

$$\mathrm{Thank}\mathrm{you}\mathrm{Sir}!$$

Commented by Rasheed.Sindhi last updated on 30/Jan/18

$$\mathrm{The}\mathrm{answer}\mathrm{in}\mathrm{the}\mathrm{book}\mathrm{is}\mathrm{\infty }.\mathrm{Certainly}$$$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{wrong}.$$

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