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Question-117688




Question Number 117688 by Khalmohmmad last updated on 13/Oct/20
Commented by prakash jain last updated on 13/Oct/20
1.  ((exponential)/(polynomial))  2. ((polynomial)/(exponential))
$$\mathrm{1}.\:\:\frac{{exponential}}{{polynomial}} \\ $$$$\mathrm{2}.\:\frac{{polynomial}}{{exponential}} \\ $$
Commented by Khalmohmmad last updated on 13/Oct/20
with is the Answer
$$\mathrm{with}\:\mathrm{is}\:\mathrm{the}\:\mathrm{Answer} \\ $$
Commented by JDamian last updated on 13/Oct/20
You should know that, when base is greater than 1, the exponential function always wins any polynomial.
Answered by 1549442205PVT last updated on 13/Oct/20
 a)  lim_(x→∞) ((4^x +2x+3)/(x^2 +6x+7))=lim_(x→∞) ((1+x.ln4+((x^2 ln^2 4)/2)+((x^3 ln^3 4)/6)++2x+3)/(x^2 +6x+7))  =lim_(x→∞) ((x^3 ln^3 4)/(6x^2 ))=∞  b)lim_(x→∞) ((x^2 +2x+3)/(3^x +3x+1))=lim((x^2 +2x+3)/(1+x.ln3+((x^2 ln^2 3)/2)+((x^3 ln^3 3)/6)+3x+1))  =lim_(x→∞)  ((6x^2 )/(x^3 ln^3 3))=0
$$\left.\:\mathrm{a}\right)\:\:\underset{\mathrm{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{4}^{\mathrm{x}} +\mathrm{2x}+\mathrm{3}}{\mathrm{x}^{\mathrm{2}} +\mathrm{6x}+\mathrm{7}}=\underset{\mathrm{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}+\mathrm{x}.\mathrm{ln4}+\frac{\mathrm{x}^{\mathrm{2}} \mathrm{ln}^{\mathrm{2}} \mathrm{4}}{\mathrm{2}}+\frac{\mathrm{x}^{\mathrm{3}} \mathrm{ln}^{\mathrm{3}} \mathrm{4}}{\mathrm{6}}++\mathrm{2x}+\mathrm{3}}{\mathrm{x}^{\mathrm{2}} +\mathrm{6x}+\mathrm{7}} \\ $$$$\underset{\mathrm{x}\rightarrow\infty} {=\mathrm{lim}}\frac{\mathrm{x}^{\mathrm{3}} \mathrm{ln}^{\mathrm{3}} \mathrm{4}}{\mathrm{6x}^{\mathrm{2}} }=\infty \\ $$$$\left.\mathrm{b}\right)\underset{\mathrm{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{2x}+\mathrm{3}}{\mathrm{3}^{\mathrm{x}} +\mathrm{3x}+\mathrm{1}}=\mathrm{lim}\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{2x}+\mathrm{3}}{\mathrm{1}+\mathrm{x}.\mathrm{ln3}+\frac{\mathrm{x}^{\mathrm{2}} \mathrm{ln}^{\mathrm{2}} \mathrm{3}}{\mathrm{2}}+\frac{\mathrm{x}^{\mathrm{3}} \mathrm{ln}^{\mathrm{3}} \mathrm{3}}{\mathrm{6}}+\mathrm{3x}+\mathrm{1}} \\ $$$$\underset{\mathrm{x}\rightarrow\infty} {=\mathrm{lim}}\:\frac{\mathrm{6x}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{3}} \mathrm{ln}^{\mathrm{3}} \mathrm{3}}=\mathrm{0} \\ $$

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