Altra was founded with a desire to find a better shoe for road and trail running. Experience the difference of our FootShape™ and ZeroDrop™ design. Welcome to Zero Limits Chiropractic located at 451 Park Fair Drive Suite 4 in Sacramento, CA. We understand that choosing a chiropractor is a big decision. This article will help you learn the basics of chiropractic care. Please don't hesitate to contact our Sacramento chiropractors with any further questions you have. We are here to help you. In Zero Limits, bestselling personal development author Craig Beck shows you how to dramatically change your perception of risk and fear, so you can easily smash right through your current self-imposed limitations. The closer (x ) gets to zero from the right the larger (in the positive sense) the function gets, while the closer (x ) gets to zero from the left the larger (in the negative sense) the function gets. Finally, the normal limit, in this case, will not exist since the two one-sided limits have different values. Subliminal Manifestation: Zero Limits uses the ancient Hawaiian healing methodology called ho'oponopono, the methodology that Dr. Hew Len used to heal an entire ward of criminally insane patients in the Hawaiian Correctional System.

Show Mobile NoticeShow All NotesHide All Notes

You appear to be on a device with a 'narrow' screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

Section 2-5 : Computing Limits

In the previous section we saw that there is a large class of functions that allows us to use

[mathop {lim }limits_{x to a} fleft( x right) = fleft( a right)]

to compute limits. However, there are also many limits for which this won’t work easily. The purpose of this section is to develop techniques for dealing with some of these limits that will not allow us to just use this fact.

Let’s first go back and take a look at one of the first limits that we looked at and compute its exact value and verify our guess for the limit.

Example 1 Evaluate the following limit. [mathop {lim }limits_{x to 2} frac{{{x^2} + 4x - 12}}{{{x^2} - 2x}}] Show Solution

First let’s notice that if we try to plug in (x = 2) we get,

[mathop {lim }limits_{x to 2} frac{{{x^2} + 4x - 12}}{{{x^2} - 2x}} = frac{0}{0}]

So, we can’t just plug in (x = 2) to evaluate the limit. So, we’re going to have to do something else.

The first thing that we should always do when evaluating limits is to simplify the function as much as possible. In this case that means factoring both the numerator and denominator. Doing this gives,

[begin{align*}mathop {lim }limits_{x to 2} frac{{{x^2} + 4x - 12}}{{{x^2} - 2x}} & = mathop {lim }limits_{x to 2} frac{{left( {x - 2} right)left( {x + 6} right)}}{{xleft( {x - 2} right)}} & = mathop {lim }limits_{x to 2} frac{{x + 6}}{x}end{align*}]

So, upon factoring we saw that we could cancel an (x - 2) from both the numerator and the denominator. Upon doing this we now have a new rational expression that we can plug (x = 2) into because we lost the division by zero problem. Therefore, the limit is,

[mathop {lim }limits_{x to 2} frac{{{x^2} + 4x - 12}}{{{x^2} - 2x}} = mathop {lim }limits_{x to 2} frac{{x + 6}}{x} = frac{8}{2} = 4]

Note that this is in fact what we guessed the limit to be.

Before leaving this example let’s discuss the fact that we couldn’t plug (x = 2) into our original limit but once we did the simplification we just plugged in (x = 2) to get the answer. At first glance this may appear to be a contradiction.

In the original limit we couldn’t plug in (x = 2) because that gave us the 0/0 situation that we couldn’t do anything with. Upon doing the simplification we can note that,

[frac{{{x^2} + 4x - 12}}{{{x^2} - 2x}} = frac{{x + 6}}{x}hspace{0.25in}{mbox{provided }}x ne 2]

In other words, the two equations give identical values except at (x = 2) and because limits are only concerned with that is going on around the point (x = 2) the limit of the two equations will be equal. More importantly, in the simplified version we get a “nice enough” equation and so what is happening around (x = 2) is identical to what is happening at (x = 2).

We can therefore take the limit of the simplified version simply by plugging in (x = 2) even though we couldn’t plug (x = 2) into the original equation and the value of the limit of the simplified equation will be the same as the limit of the original equation.

On a side note, the 0/0 we initially got in the previous example is called an indeterminate form. This means that we don’t really know what it will be until we do some more work. Typically, zero in the denominator means it’s undefined. However, that will only be true if the numerator isn’t also zero. Also, zero in the numerator usually means that the fraction is zero, unless the denominator is also zero. Likewise, anything divided by itself is 1, unless we’re talking about zero.

So, there are really three competing “rules” here and it’s not clear which one will win out. It’s also possible that none of them will win out and we will get something totally different from undefined, zero, or one. We might, for instance, get a value of 4 out of this, to pick a number completely at random.

When simply evaluating an equation 0/0 is undefined. However, in take the limit, if we get 0/0 we can get a variety of answers and the only way to know which on is correct is to actually compute the limit.

There are many more kinds of indeterminate forms and we will be discussing indeterminate forms at length in the next chapter.

Let’s take a look at a couple of more examples.

Example 2 Evaluate the following limit. [mathop {lim }limits_{h to 0} frac{{2{{left( { - 3 + h} right)}^2} - 18}}{h}] Zero limits 2

Show Solution

In this case we also get 0/0 and factoring is not really an option. However, there is still some simplification that we can do.

[begin{align*}mathop {lim }limits_{h to 0} frac{{2{{left( { - 3 + h} right)}^2} - 18}}{h} & = mathop {lim }limits_{h to 0} frac{{2left( {9 - 6h + {h^2}} right) - 18}}{h} & = mathop {lim }limits_{h to 0} frac{{18 - 12h + 2{h^2} - 18}}{h} & = mathop {lim }limits_{h to 0} frac{{ - 12h + 2{h^2}}}{h}end{align*}]

So, upon multiplying out the first term we get a little cancellation and now notice that we can factor an (h) out of both terms in the numerator which will cancel against the (h) in the denominator and the division by zero problem goes away and we can then evaluate the limit.

[begin{align*}mathop {lim }limits_{h to 0} frac{{2{{left( { - 3 + h} right)}^2} - 18}}{h} & = mathop {lim }limits_{h to 0} frac{{ - 12h + 2{h^2}}}{h} & = mathop {lim }limits_{h to 0} frac{{hleft( { - 12 + 2h} right)}}{h} & = mathop {lim }limits_{h to 0} , - 12 + 2h = - 12end{align*}] Example 3 Evaluate the following limit. [mathop {lim }limits_{t to 4} frac{{t - sqrt {3t + 4} }}{{4 - t}}] Show Solution

This limit is going to be a little more work than the previous two. Windows xp genuine update. Once again however note that we get the indeterminate form 0/0 if we try to just evaluate the limit. Also note that neither of the two examples will be of any help here, at least initially. We can’t factor the equation and we can’t just multiply something out to get the equation to simplify.

When there is a square root in the numerator or denominator we can try to rationalize and see if that helps. Recall that rationalizing makes use of the fact that

[left( {a + b} right)left( {a - b} right) = {a^2} - {b^2}]

So, if either the first and/or the second term have a square root in them the rationalizing will eliminate the root(s). This might help in evaluating the limit.

Let’s try rationalizing the numerator in this case.

[mathop {lim }limits_{t to 4} frac{{t - sqrt {3t + 4} }}{{4 - t}} = mathop {lim }limits_{t to 4} frac{{left( {t - sqrt {3t + 4} } right)}}{{left( {4 - t} right)}},frac{{left( {t + sqrt {3t + 4} } right)}}{{left( {t + sqrt {3t + 4} } right)}}]

Remember that to rationalize we just take the numerator (since that’s what we’re rationalizing), change the sign on the second term and multiply the numerator and denominator by this new t