prove the limit using definition.

Question

1) $\lim_{(n) \rightarrow (\infty)} \frac{4n+7}{2n-4}= 2$

2) $\lim_{(n) \rightarrow (\infty)} \frac{7n^2+9n-17}{4n^2-5n+6}= 7/4$

in first question I got that $ \frac{15}{|2n-4|}$ < $\epsilon$ , how can I get red of the absolute value in this case?

Answer 1

$\frac{15}{|2n-4|} < \epsilon \rightarrow |2n-4| > \frac{15}{\epsilon}$ So...

$2n-4 > \frac{15}{\epsilon}$ or $2n-4 < \frac{15}{-\epsilon}$. The second doesn't make sense, so choose $2n-4 > \frac{15}{\epsilon}$ which yields

$n > \frac{\frac{15}{\epsilon} + 4}{2}$

Answer 2

From the definition $$ \left|\frac{4n+7}{2n-4}-2 \right|<\varepsilon $$ for large $n$. But $$ \left|\frac{4n+7}{2n-4}-2 \right|=\left| \frac{15}{2n-4} \right|. $$ However, $n$ is natural, hence it is equal to $\dfrac{15}{2n-4}$.

Hence (an exemplary) answer is $$ n_0=\max\left(3,\left\lfloor\frac12\left(\frac{15}{\varepsilon}+4\right) \right\rfloor\right). $$

Answer 3

I will do the first one, the easy one (numer (1)), and leave the other one, the harder one for you:

Let $\epsilon > 0 $ be given. We want to find a positive integer $N$ such that

$$ | \frac{4n + 7}{2n-4} - 2 | < \epsilon \; \; ....................(I)$$

when $n > N$. Note: (for $n > 2 $)

$$ | \frac{4n + 7}{2n-4} - 2 | = | \frac{15}{2n-4} | = \frac{15}{2n-4} $$

$$ \frac{15 }{2n - 4} < \epsilon \iff \frac{15}{\epsilon}+4 < 2n \iff \frac{15 \epsilon + 4}{2 \epsilon} < n$$

Hence, you want to choose $N$ to be the integer part of $ \frac{ 15 \epsilon + 4 }{ 2 \epsilon } $. With this choice, it is clear that $(I)$ is satisfied when $n > N$