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Question

A positive number

ϵ\epsilon

and the limit L of a function f at

++ \infty

are given. Find a positive number N such that |f(x)-L|<

ϵ\epsilon

if x>N.

limx+4x12x+5=2;ϵ=0.1\lim _ { x \rightarrow + \infty } \frac { 4 x - 1 } { 2 x + 5 } = 2 ; \epsilon = 0.1

Solution

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Answered 1 year ago
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44.\boxed{44.} limx+4x12x+5=2\lim\limits_{x \to +{\infty}}\dfrac{4x-1}{2x+5}=2 ; ϵ=0.1\epsilon=0.1 \

We know that this limit exists and is equal to 2, so using Definition 1.4.2\text{\color{#c34632}Definition 1.4.2} we know that for ϵ=0.1\epsilon=0.1 we can choose a positive integer NN such that inequality 4x12x+52<ϵ\left| \dfrac{4x-1}{2x+5}-2 \right|<\epsilon is satisfied for all x>Nx> N.

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