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How do I evaluate this limit using the definition?

$$\lim_{x \rightarrow \infty} \frac{2x+3}{x-4}$$

I know it equals 2, but the delta-epsilon argument trips me up here.

Thanks in advance.

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A related problem. – Mhenni Benghorbal Mar 20 '13 at 12:28
up vote 4 down vote accepted

$$ f(x) = \frac{2x+3}{x-4} = 2 + \frac{11}{x-4}. $$ That much follows from long division: the quotient is $2$ and the remainder is $11$. You want $f(x)$ to differ from $2$ by less than $\varepsilon>0$ when $x$ is big enough. The question is how big is big enough. The difference between $f(x)$ and $2$ is the second term above, $11/(x-4)$. Since $x-4$ is positive, you can say $11/(x-4)<\varepsilon$ if $x>\frac{11}{\varepsilon}+4$.

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so does this $\frac{11}{\epsilon} + 4$ become my $d$? – Peej Gerard Mar 20 '13 at 12:45
Depends on what you mean by $d$. If you're tyring to prove that for every $\varepsilon>0$, there exists $d$ such that $|f(x)-2|<\varepsilon$ whenever $x>d$, then you can use $d=\frac{11}{\varepsilon}+4$. – Michael Hardy Mar 20 '13 at 12:47
that is indeed what I mean, sorry. used the notion I was given in class without being clear! thanks again. – Peej Gerard Mar 20 '13 at 12:48

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