Prove a limit equality

[vast phoenix]

Prove that
lim_n->∞ (4n^2 + 1)/(3n^2 + 2) = 4/3
by using the epsilon delta definition

[chilly kite]

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[queen abyss]

Why do you need the epsilon-delta proof?

Although one could arguably say that:
$$\frac{4n^2+1}{3n^2+1} = \frac{4}{3} \times \frac{1 + \frac{1}{4n^2}}{1 + \frac{2}{3n^2}}$$

[indigo sand]

Rion

[queen abyss]

So taking the candidate limit L:
$$\frac{4n^2+1}{3n^2+1} - L = \frac{4}{3} \left( \frac{1 + \frac{1}{4n^2}}{1 + \frac{2}{3n^2}}-1 \right)$$

[indigo sand]

Rion

[queen abyss]

I think you can work your way from there with some inequality tricks

I'll let you try out

[vast phoenix]

Can you explain your reasoning more thoroughly? I don't get it

What is the candidate limit and why are you subtracting it?

The problem has a constraint to use the definition

[queen abyss]

4/3

that's what you said in the question

[vast phoenix]

I get it

And why are you subtracting it?

[queen abyss]

Definition of the limit you were looking for

[hearty kestrel]

Do you know the epsilon delta definition?

[vast phoenix]

f(x) tends to L as x -> c iff For any epsilon > 0 there is some delta > 0 such that for any x (0 < |x - c| < delta => |f(x) - L| < epsilon)

Is that correct?