Cross your limits 3

Calculus Level 4

lim x e x ( ( 2 x n ) 1 / e x ( 3 x n ) 1 / e x ) x n \large \lim_{x\to\infty} \dfrac{e^x \left( \left(2^{x^n}\right)^{1/e^x} - \left(3^{x^n}\right)^{1/e^x} \right)}{x^n}

Evaluate the limit above for positive integer n n .

None of these choices 0 0 ln 2 ln 3 \ln2 - \ln 3 ln 3 ln 2 \ln3 - \ln 2

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1 solution

Rudraksh Shukla
May 1, 2016

I will use evaluation using limit by substitution, Substitute x n e x \dfrac{x^n} {e^x} as t t . Now as x x\rightarrow \infty , x n e x 0 \dfrac{x^n}{e^x} \rightarrow 0 (I am not sure about my proof for this but I think it holds true).

So we shall have the question as, lim t 0 2 t 3 t t \lim_{t\to{0}} \dfrac{2^t-3^t}{t} Add and subtract 1, lim t 0 2 t 1 ( 3 t 1 ) t \lim_{t\to{0}} \dfrac{2^t-1-(3^t-1)}{t} = lim t 0 2 t 1 t lim t 0 3 t 1 t =\lim_{t\to{0}} \dfrac{2^t-1}{t} - \lim_{t\to{0}} \dfrac{3^t-1}{t} Using the relation, lim f ( x ) 0 a f ( x ) 1 f ( x ) = l n ( a ) \lim_{f(x)\to{0}} \dfrac{a^{f(x)}-1}{f(x)}=ln(a) The expression is equal to l n ( 2 ) l n ( 3 ) ln(2)-ln(3)

For completeness, you should prove that lim x x n e x = 0 \displaystyle \lim_{x \to \infty} \dfrac{x^n}{e^x} =0 .

Here's the proof though:

Let x > 0 x>0

Then, e x = r = 0 x r r ! e x > x n + 1 ( n + 1 ) ! > 0 ( n + 1 ) ! x > x n e x > 0 0 = lim x ( n + 1 ) ! x lim x x n e x 0 \begin{aligned} e^x&=\sum_{r=0}^{\infty} \frac{x^r}{r!} \\\implies e^x&>\frac{x^{n+1}}{(n+1)!}>0 \\\implies \frac{(n+1)!}{x}&>\frac{x^n}{e^x}>0 \\\implies 0=\lim_{x \to \infty}\frac{(n+1)!}{x}&\geq \lim_{x \to \infty} \frac{x^n}{e^x}\geq 0 \end{aligned}

Hence, by using squeeze theorem , the proof is complete.

@Rudraksh Shukla

A Former Brilliant Member - 5 years, 1 month ago

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Thank you for the proof!

Rudraksh Shukla - 5 years, 1 month ago

it is absolutely correct. for using substitutions in limits you just have to ensure that the quantity you are substituting is the same the the numerator and debominator. or you can check that the individual derivative of the quantity cancels out from the numerator and denominator while applying L Hospital.

Antara Chatterjee - 3 years ago

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