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Question Number 80296 by john santu last updated on 02/Feb/20
what is the value of   lim_(x→−∞ )  e^((6x^2 +x)/(3x+5))  ?  0 or ∞ ?
$${what}\:{is}\:{the}\:{value}\:{of}\: \\ $$$$\underset{{x}\rightarrow−\infty\:} {\mathrm{lim}}\:{e}^{\frac{\mathrm{6}{x}^{\mathrm{2}} +{x}}{\mathrm{3}{x}+\mathrm{5}}} \:? \\ $$$$\mathrm{0}\:{or}\:\infty\:? \\ $$
Commented by Tony Lin last updated on 02/Feb/20
let t=−x  lim_(t→+∞) e^((6t^2 −t)/(5−3t))   =e^(lim_(t→+∞) ((6t^2 −t)/(5−3t)))    lim_(t→+∞) ((6t^2 −t)/(5−3t))=lim_(t→+∞) ((12t−1)/(−3))=−∞  e^(−∞) =0
$${let}\:{t}=−{x} \\ $$$$\underset{{t}\rightarrow+\infty} {\mathrm{lim}}{e}^{\frac{\mathrm{6}{t}^{\mathrm{2}} −{t}}{\mathrm{5}−\mathrm{3}{t}}} \\ $$$$={e}^{\underset{{t}\rightarrow+\infty} {\mathrm{lim}}\frac{\mathrm{6}{t}^{\mathrm{2}} −{t}}{\mathrm{5}−\mathrm{3}{t}}} \: \\ $$$$\underset{{t}\rightarrow+\infty} {\mathrm{lim}}\frac{\mathrm{6}{t}^{\mathrm{2}} −{t}}{\mathrm{5}−\mathrm{3}{t}}=\underset{{t}\rightarrow+\infty} {\mathrm{lim}}\frac{\mathrm{12}{t}−\mathrm{1}}{−\mathrm{3}}=−\infty \\ $$$${e}^{−\infty} =\mathrm{0} \\ $$
Commented by jagoll last updated on 02/Feb/20
Commented by jagoll last updated on 02/Feb/20
good
$${good} \\ $$

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