math proofs: use the definition of the limit of a sequence

Question

I am asked to use the definition of the limit of a sequence to prove: $$\lim_{n\to\infty}\frac{3n-1}{2n+5}=\frac32.$$ My hint: this is only requires creative choosing of $N$ and applying the definition. So thus far,I have: $n\ge N \implies \left|\frac{3n-1}{2n+5}.\frac32\right|< \epsilon$.

Anyway, I figured out I need solve for little $n$. However each time I do that, I end up with a negative value of $-\frac{17}{4}\epsilon-\frac52$ and then I just get stuck and don't know what to do. Could anyone please be of help? Thanks.

Answer 1

Hint: $\frac{3n-1}{2n+5}=\frac{\left(3n+\frac{15}2\right)-\frac{17}2}{2n+5}$

Answer 2

Intuitively as $n\to\infty$, the terms $an$ and $cn$ will be dominant in the expression $\frac{an+b}{cn+d}$ since they are increasing linearly whereas the terms $b$ and $d$ are constant. We should expect $\frac{an+b}{cn+d}\sim\frac{an}{cn}=\frac{a}{c}$.

This means the numerator should be "about" the denominator times $\frac{a}{c}$. In fact:

$$an+b=\frac{a}{c}(cn+d)+\square \tag{$\circ$}$$

for some constant $\square$. So we have

$$\frac{an+b}{cn+d}-\frac{a}{c}=\frac{\frac{a}{c}(cn+d)+\square}{cn+d}-\frac{a}{c}=\frac{\square}{cn+d}.$$

Given an $\varepsilon>0$, how big does $n$ have to be to guarantee $|\frac{\square}{cn+d}|<\varepsilon$?

(Note you can solve for $\square$ explicitly in $(\circ)$. Use your numbers for $a,b,c,d$ if you wish.)

Answer 3

You are doing everything fine, except you're dropping the absolute values:

Since $\;\displaystyle\left|\frac{3n-1}{2n+5}-\frac{3}{2}\right|=\left|\frac{-17}{2(2n+5)}\right|=\frac{17}{2(2n+5)}$, you just have to solve $\displaystyle\frac{17}{2(2n+5)}<\epsilon$

to find a value of $N$ that will work.

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You could also use $\displaystyle\frac{17}{2(2n+5)}<\frac{20}{2(2n+5)}=\frac{10}{2n+5}<\frac{10}{2n}=\frac{5}{n}$

and then solve $\displaystyle\frac{5}{n}<\epsilon$ to find a simpler expression for $N$,$\;\;$if you wanted to.