Limit problem

Calculus Level 4

lim x ( 34 2 x + 34 3 x ) 1 x = ? \large\lim_{x\to \infty} \left(342^x+343^x\right)^{\frac{1}{x}} = \, ?


The answer is 343.

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1 solution

This limit can be written as lim x 343 × ( 1 + ( 342 343 ) x ) 1 x = 343 \displaystyle\lim_{x \to \infty} 343 \times \left(1 + \left(\dfrac{342}{343}\right)^{x}\right)^{\frac{1}{x}} = \boxed{343} ,

since ( 1 + a x ) 1 x (1 + a^{x})^{\frac{1}{x}} for 0 < a < 1 0 \lt a \lt 1 is of the form 1 0 = 1 1^{0} = 1 as x x \to \infty .

Comment: In general, for 0 < a b 0 \lt a \le b , lim x ( a x + b x ) 1 x = b \displaystyle\lim_{x \to \infty} (a^{x} + b^{x})^{\frac{1}{x}} = b .

Can you prove that lim x ( 34 2 x + 34 3 x ) 1 x = lim x 343 × ( 1 + ( 342 343 ) x ) 1 x \lim_{x\to \infty} \left(342^x+343^x\right)^{\frac{1}{x}} = \lim_{x\to \infty} 343 \times \left(1+ \left(\frac{342}{343} \right)^x\right)^{\frac{1}{x}}

And why is ( 1 + a x ) 1 x (1+ a^x)^{\frac{1}{x}} for 0 < a < 1 0 < a < 1 is of the form 1 0 = 1 1^0 = 1 as x x \to \infty ? The limit of a x a^x as x x \to \infty is 0 0 ?

Jason Chrysoprase - 4 years, 4 months ago

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For the first question, we have that

( 34 2 x + 34 3 x ) 1 x = ( 34 3 x ( 1 + ( 342 343 ) x ) ) 1 x = (342^{x} + 343^{x})^{\frac{1}{x}} = \left(343^{x}\left(1 + \left(\dfrac{342}{343}\right)^{x}\right)\right)^{\frac{1}{x}} =

343 × ( 1 + ( 342 343 ) x ) 1 x 343 \times \left(1 + \left(\dfrac{342}{343}\right)^{x}\right)^{\frac{1}{x}} ,

since ( 34 3 x ) 1 x = 343 (343^{x})^{\frac{1}{x}} = 343 . So if the first expression has a limit then it will be the same as that for the last expression.

For the second question, yes, since a x 0 a^{x} \to 0 for 0 < a < 1 0 \lt a \lt 1 as x x \to \infty we have ( 1 + a x ) 1 x (1 + a^{x})^{\frac{1}{x}} being of the form 1 0 1^{0} , which is not one of the indeterminate forms , so we can go ahead and conclude that the limit is 1 0 = 1 1^{0} = 1 .

We can also look at it this way. If y = ( 1 + a x ) 1 x y = (1 + a^{x})^{\frac{1}{x}} then ln ( y ) = ln ( 1 + a x ) x \ln(y) = \dfrac{\ln(1 + a^{x})}{x} , which as x x \to \infty goes to ln ( 1 ) x 0 \dfrac{\ln(1)}{x} \to 0 , and thus y e 0 = 1 y \to e^{0} = 1 .

Brian Charlesworth - 4 years, 4 months ago

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I see, thank you

Jason Chrysoprase - 4 years, 4 months ago

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