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  • Derivatives: Chain Rule
  • Derivatives: Definition of a Derivative
  • Derivatives: Product and Quotient Rule
  • Derivatives: Properties of, Power Rule
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  • Integrals: Indefinite Integrals, Power Rule
  • Integrals: Properties of Definite Integrals
  • Integrals: Substitution
  • Limits
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  • Vectors, continued

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Limits

Limits are useful in a variety of areas, and especially in calculus.  The standard definition of the derivative uses limits to evaluate the derivative of a function.  In this section we will cover two definitions of a limit, and then go through some examples.

Definition of a limit

The limit of f(x) is L as x approaches a is written as:

where we can make f(x) as close to L as we like for all x which are close to a.

More precise definition of a limit

For a function f(x) that contains values near x=a, the limit

if for every \epsilon \, >\, 0 there is some \delta \, > \, 0 such that

\left| \, f(x)-L \, \right| \, < \, \epsilon     whenever     0 \, < \, \left| x-a \, \right| \, < \, \delta

The second definition is a little more intimidating, but can be understood with some work.  With the definitions out of the way, sometimes it is best to see some examples first before gaining an greater understanding of the definition

Properties of limits

Examples of limits

First we will start with some basic examples, and finish up with some harder problems.

1.\,\, \lim_{x\to2} \, \frac{1}{x+1}=\frac{1}{3}

As you can see, this simple example was done by plugging in x is 2, for x into 1/(x+1), which gives us 1/3.  What it is actually saying is, that as x approaches the value of 2, 1/(x+1) approaches 1/3

 

2.\,\, \lim_{x\to1}\,\frac{x-1}{x^2+x+1} = \frac{1-1}{1^2+1+1}=\frac{0}{3}=0

Once again, this example is quite simple, just plug in the value of x=1.

 

3.\,\, \lim_{x\to2}\, \frac{x-2}{(x-2)(x+4)} = \frac{2-2}{(2-2)(2+4)}=\frac{0}{0}

In this example we have a limit which is approaching 0/0 -- an indefinite value.  This is what mathematicians call a 'hole'.  That is because the function is undefined at f(2), but still approaches that value from the left and right sides.  In this case, you have to reduce the fraction to evaluate the limit since,

\frac{x-2}{(x-2)(x+4)} = \frac{1}{x+4}

then,

\lim_{x\to2} \, \frac{1}{x+4}=\frac{1}{2+4}=\frac{1}{6}

hence, as x approaches 2, f(x) approaches \frac{1}{6}.  However, the value of x=2 is undefined, so the limit only approaches this value, coming very close to it, but the function itself is not defined at x=2.

 

 4. \,\, \lim_{x\to0} \, \frac{1}{x} = \frac{1}{0} = \infty

A relatively simple example, showing that 1/0 is infinity.  So the function 1/x approaches infinity as x approaches 0.

 

 5. \,\, \lim_{x\to\infty} \, \frac{x^2+x+1}{x^3+5x^2+3x+5}

How do we approach such a problem?  Clearly, if you just 'plug in' \infty into x you will get \frac{\infty}{\infty}, an indeterminate value.  These type of problems can be done using the following technique:

Start by dividing the top and bottom of the fraction by the greatest degree term that occurs in the rational expression:

 \frac{(x^2+x+1)\frac{1}{x^3}}{(x^3+5x^2+3x+5)\frac{1}{x^3}}

multiplying these terms through gives:

 \frac{\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}}{1+\frac{5}{x}+\frac{3}{x^2}+\frac{5}{x^3}}

Now we are in a position to evaluate the limit, since most terms in the numerator and denominator go to 0 as x goes to \infty

 \lim_{x\to\infty} \, \frac{\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}}{1+\frac{5}{x}+\frac{3}{x^2}+\frac{5}{x^3}}=\frac{0+0+0}{1+0+0+0}=\frac{0}{1}=0

Perhaps some initial reasoning could have confirmed that the limit approached 0, however it is also instructive to show your work in this type of problem, since not all of these will approach 0 as in the next example.

6.\,\, \lim_{x\to\infty} \, \frac{2x^2+5x+1}{3x^2+500x+3}

Although in problem number 5, it was perhaps easy to see that the limit would approach 0 as x went to infinity, in this problem it isn't apparent what the limit approaches (unless of course you are well practiced in these type of problems).  However we can use the same method as in problem 5 to evaluate this limit:

 \frac{2x^2+5x+1}{3x^2+500x+3} = \frac{(2x^2+5x+1)\frac{1}{x^2}}{(3x^2+500x+3)\frac{1}{x^2}}=\frac{2+\frac{5}{x}+\frac{1}{x^2}}{3+\frac{500}{x}+\frac{3}{x^2}}

We are now in a position to evaluate this limit,

 \lim{x\to\infty} \frac{2+\frac{5}{x}+\frac{1}{x^2}}{3+\frac{500}{x}+\frac{3}{x^2}}=\frac{2+0+0}{3+0+0} = \frac{2}{3}

Hence the limit approaches \frac{2}{3} as x approaches \infty

 7.

Are you still with us?  Good!  In this last example we will show how to use the more precise definition of a limit.

Use the precise definition of the limit, to prove that

 \lim_{x\to1}\, 4x+3=7

Once again, the definition of the limit goes as follows:

 

For a function f(x) that contains values near x=a, the limit

if for every \epsilon \, >\, 0 there is some \delta \, > \, 0 such that

\left| \, f(x)-L \, \right| \, < \, \epsilon     whenever     0 \, < \, \left| x-a \, \right| \, < \,  \delta

 

for our problem, we have a = 1 , L = 7, and f(x)=4x+3.  \epsilon and \delta are some arbitrary numbers, which are useful in the proof.  So according to the definition of the limit we need the following to be true:

\left| \, 4x+3-7 \, \right| \, < \, \epsilon     whenever     0 \, < \, \left| x-1 \, \right| \, < \,   \delta

and after some simplification:

\left| \, 4x-4 \, \right| \, < \, \epsilon     whenever     0 \, < \, \left| x-1 \, \right| \, < \,    \delta

to move on in this proof, we want to make the inequality using \epsilon to look like the inequality which uses {tex]\delta{/tex}.  Factoring out 4, and then dividing both sides by 4 should do the trick.

\left| \, x-1 \, \right| \, < \, \frac{\epsilon}{4}     whenever     0 \, < \, \left| x-1 \, \right| \, < \,     \delta

These two inequalities are now looking very similar, this leads us to choose the guess that

\delta=\frac{\epsilon}{4}

We will now see if our guess can be verified.  For \epsilon\,>\,0 choose \delta=\frac{\epsilon}{5}

Now we assume that 0\,< \left|x-1\right| \, < \delta = \frac{\epsilon}{4} and get:

\left|\,4x+3-7\,\right| = \left|\,4x-4\,\right| = 4\left|\,x-1\,\right| = 4(\frac{\epsilon}{4})={\epsilon}

Therefore, we have shown that,

\left| \, 4x+3-7 \, \right| \, < \, \epsilon     whenever     0 \, < \, \left| x-1 \, \right| \, < \, \frac{\epsilon}{4}

and by definition,

 \lim_{x\to1}\, 4x+3=7

 
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