# Limits using definition

• Aug 17th 2007, 12:00 PM
Dili
Limits using definition
Can you tell me how to find a limit using the definition only?
For example
Limit of 1/(5x-3) when x tends to 2.

I do know how to prove the above limit is 1/7(if the value 1/7 is given) using the definition. But i do not know how to arrive at the value using the definition of limits

Thanks
• Aug 17th 2007, 12:43 PM
topsquark
Quote:

Originally Posted by Dili
Can you tell me how to find a limit using the definition only?
For example
Limit of 1/(5x-3) when x tends to 2.

I do know how to prove the above limit is 1/7(if the value 1/7 is given) using the definition. But i do not know how to arrive at the value using the definition of limits

Thanks

Is this what you are looking for?

$\displaystyle \lim_{x \to 2^-}\frac{1}{5x - 3} = \frac{1}{7}$
and
$\displaystyle \lim_{x \to 2^+}\frac{1}{5x - 3} = \frac{1}{7}$

Since the limit exists on both sides and is the same we may unambigously state that
$\displaystyle \lim_{x \to 2}\frac{1}{5x - 3} = \frac{1}{7}$

If this isn't what you are looking for, I'm not sure what to say. The point x = 2 is in the domain of the expression, and as such we can usually say that the limit is just the value of the expression with x = 2. (Because of the argument above.)

-Dan
• Aug 17th 2007, 12:44 PM
topsquark
Oh! Unless you mean you need an $\displaystyle \epsilon - \delta$ proof? (I'd have to look that one up myself to make sure I've got the definitions right.)

-Dan
• Aug 17th 2007, 01:31 PM
Plato
This is just an exercise in algebra.
Note that: $\displaystyle \left| {\frac{1}{7} - \frac{1}{{5x - 3}}} \right| = \left| {\frac{{5x - 10}}{{7\left( {5x - 3} \right)}}} \right|$.

If $\displaystyle \left| {x - 2} \right| < 1$ then $\displaystyle \left| {\frac{1}{{7\left( {5x - 3} \right)}}} \right| < \frac{1}{{14}}$

Here it is. If $\displaystyle \varepsilon > 0$ then choose $\displaystyle \delta = \min \left\{ {\frac{\varepsilon }{2},1} \right\}$, then if $\displaystyle 0 < \left| {x - 2} \right| < \delta$ we have $\displaystyle \left| {\frac{{5x - 10}}{{7\left( {5x - 3} \right)}}} \right| = \frac{{5\left| {x - 2} \right|}}{{\left| {7\left( {5x - 3} \right)} \right|}} \le \frac{5}{{14}}\delta < \varepsilon$.

QED
• Aug 17th 2007, 08:31 PM
Dili
Yes, by definition i meant the epsilon delta proof.
But I do know how to prove lim 1/(5x-3) is 1/7 when x tends to 2.(the way Plato proved)
The problem is we were given the limit, just to find the limit of 1/(5x-3) when x tends to 2 using the definition. We were not given the value 1/7, we have to arrive at it
If anyone can help i would be really grateful

Thanks
• Aug 18th 2007, 03:36 AM
Plato
Quote:

Originally Posted by Dili
The problem is we were given the limit, just to find the limit of 1/(5x-3) when x tends to 2 using the definition. We were not given the value 1/7, we have to arrive at it.

I think that you have misunderstood the question.
There really is no applying the $\displaystyle \varepsilon \& \delta$ method to find the limit.
We use the definition to prove that 1/7 is the limit once we have found it by usual means.