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Question Number 116859 by Study last updated on 07/Oct/20

lim_(x→∞) ((3x^2 )/5^x )=?

$${li}\underset{{x}\rightarrow\infty} {{m}}\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{5}^{{x}} }=? \\ $$

Commented by JDamian last updated on 07/Oct/20

As exponential function increases more rapidly than parabola, you even can get this limit intuitively

Answered by bemath last updated on 07/Oct/20

lim_(x→∞)  ((6x)/(5^x .ln (5))) = lim_(x→∞)  (6/(5^x .(ln (5))^2 )) = 0

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{6x}}{\mathrm{5}^{\mathrm{x}} .\mathrm{ln}\:\left(\mathrm{5}\right)}\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{6}}{\mathrm{5}^{\mathrm{x}} .\left(\mathrm{ln}\:\left(\mathrm{5}\right)\right)^{\mathrm{2}} }\:=\:\mathrm{0} \\ $$

Commented by bemath last updated on 07/Oct/20

Answered by 1549442205PVT last updated on 07/Oct/20

lim_(x→∞) ((3x^2 )/5^x )= This is the form (∞/∞).Hence,  using L′Hopital rule we have  I=lim_(x→∞) ((3x^2 )/5^x )=lim_(x→∞) ((6x)/(5^x ln5))=lim_(x→∞) (6/(5^x (ln5)^2 ))  =0

$${li}\underset{{x}\rightarrow\infty} {{m}}\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{5}^{{x}} }=\:\mathrm{This}\:\mathrm{is}\:\mathrm{the}\:\mathrm{form}\:\frac{\infty}{\infty}.\mathrm{Hence}, \\ $$$$\mathrm{using}\:\mathrm{L}'\mathrm{Hopital}\:\mathrm{rule}\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{I}={li}\underset{{x}\rightarrow\infty} {{m}}\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{5}^{{x}} }=\underset{\mathrm{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{6x}}{\mathrm{5}^{\mathrm{x}} \mathrm{ln5}}=\underset{\mathrm{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{6}}{\mathrm{5}^{\mathrm{x}} \left(\mathrm{ln5}\right)^{\mathrm{2}} } \\ $$$$=\mathrm{0} \\ $$

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