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So far have this;

Need to show $\forall \epsilon >0 \; \exists N=N(\epsilon) \text{ s.t. } \forall n>N \; \Bigl{|}\frac{a \cdot n + b}{c \cdot n+d} - \frac{a}{c}\Bigr{|}<\epsilon \\$

So $|\frac{a \cdot n + b}{c \cdot n+d} - \frac{a}{c}| = |\frac{acn+bc-acn-ad}{c^2n+cd} - \frac{a}{c}| =| \frac{bc-ad}{c^2n+cd}| < \frac{b}{c^2n}=\frac{b}{cn} < \epsilon$

So a suitable $N \; \forall n>N=\frac{b}{c\epsilon}$

I have a few questions. Is the above correct ? Should I state assumptions about $a,b,d$ ? And finally how do I show that the limit is in fact $\frac{a}{c}$?

alortimor
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    In $| \frac{bc-ab}{c^2n+cd}| < \frac{b}{cn}$, you are assuming both $ab\geq0$ and $cd\geq0$ neither of which is justified. (I removed the extra factor of $c$ in the denominator, which is just a typo.) Also, the $-ab$ should be $-ad$ – saulspatz Apr 08 '20 at 14:08

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If you were just computing the limit, you would say $$\frac{na+b}{nc+d}=\frac{a+\frac bn}{c+\frac dn}$$ and argue that $\frac bn\to0$ and $\frac dn\to0$. I think that's the best way to approach the proof, too.

Let $\varepsilon>0$ be given and choose $N$ so large that $n>N$ implies $\left|\frac bn\right|<\varepsilon$ and $\left|\frac dn\right|<\varepsilon$. Then $$\left|\frac{an+b}{cn+d}-\frac ac\right|=\left|\frac{\frac bn-\frac{ad}{cn}}{c+\frac dn}\right|<\frac{\varepsilon+\left|\frac ac\right|\varepsilon}{|c|-\varepsilon}$$

Now we run into a problem in bounding the denominator, and we see that we should also required $\varepsilon <\frac{|c|}2$. Then the last expression is less than $$\varepsilon\frac{2+2\left|\frac ac\right|}{|c|}$$

Since this is a constant times $\varepsilon$ is can be made arbitrarily small and we are done. If you want to make the bound come out to exactly $\varepsilon$, you can let $$\varepsilon_1=\min\left\{\frac{|c|}2, \frac{|c|}{2+2\left|\frac ac\right|}\varepsilon\right\}$$ and carry out the argument with $\varepsilon_1$ instead of $\varepsilon$.

I leave to you to arrange this into a well-stated proof.

saulspatz
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