Scary Limits!!

Calculus Level 4

$\beta$ = $\displaystyle \lim_{ x \rightarrow \infty} \dfrac{(2^{x^n})^{{1}/{e^x}} -(3^{x^n})^{{1}/{e^x}} }{x^n}$

Given That n $\epsilon$ N.

Find $2^{\beta} + 2^{\beta +1}$ .

The answer is 3.

2 solutions

汶汶樂
Feb 5, 2015

Good.

$\beta = \displaystyle lim_{x \to \infty} \dfrac{e^{\frac{x^n}{e^x}log2} - e^{\frac{x^n}{e^x}log3}}{x^n}$

$\beta = \displaystyle lim_{x \to \infty} \dfrac{e^{\frac{x^n}{e^x}log2} - 1 - ( e^{\frac{x^n}{e^x}log3} - 1)}{x^n}$

$\beta = \displaystyle lim_{x \to \infty} \dfrac{e^{\frac{x^n}{e^x}log2} - 1}{x^n} - \dfrac{ e^{\frac{x^n}{e^x}log3} - 1}{x^n}$

$\beta = \displaystyle lim_{x \to \infty} \dfrac{e^{\frac{x^n}{e^x}log2} - 1}{\dfrac{x^ne^x}{log2}\times \dfrac{log2}{e^x}} - \dfrac{ e^{\frac{x^n}{e^x}log3} - 1}{\dfrac{x^ne^x}{log3}\times \dfrac{log3}{e^x}}$

$\beta =0$

U Z - 6 years, 4 months ago

Incredible Mind
Feb 5, 2015

u can do this in ur head...

lim x -> inf x^n/e^x = 0

hence ans becomes 2^0-3^0 / inf^n at x=inf.so beta is 0*0=0..

ANS is 2^0+2^1=3

Scary Limits!!

The answer is 3.

2 solutions

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