properties of limits

Properties

Limits have a number of properties that you can use to your advantage:

  1. x→alim⁡k=k ^{\lim}_{x \to a} k = k, for any constant kk

  2. x→alim⁡x=x ^{\lim}_{x \to a} x = x

  3. x→alim⁡[f(x) ± h(x)]= x→alim⁡f(x) ± x→alim⁡h(x) ^{\lim}_{x \to a} [ f(x) \ \pm \ h(x) ] = \ ^{\lim}_{x \to a} f(x) \ \pm \ ^{\lim}_{x \to a} h(x)

  4. x→alim⁡[kf(x)]=k[x→alim⁡f(x)] ^{\lim}_{x \to a} [ kf(x)] = k [ ^{\lim}_{x \to a} f(x) ], for any constant kk

  5. x→alim⁡[f(x)g(x)]=[x→alim⁡f(x)][x→alim⁡g(x)] ^{\lim}_{x \to a} [ f(x)g(x)] = [ ^{\lim}_{x \to a} f(x) ][ ^{\lim}_{x \to a} g(x) ]

  6. x→alim⁡f(x)g(x)=x→alim⁡f(x)x→alim⁡g(x) ^{\lim}_{x \to a} \frac{f(x)}{g(x)} = \frac{ ^{\lim}_{x \to a} f(x)}{ ^{\lim}_{x \to a} g(x)}, as long as x→alim⁡g(x)≠0 ^{\lim}_{x \to a} g(x) \ne 0

  7. x→alim⁡[f(x)]n=[x→alim⁡f(x)]n ^{\lim}_{x \to a} [ f(x)]^{n} = [ ^{\lim}_{x \to a} f(x)]^{n}, for any rational number nn

Most of these are pretty easy to remember, however, the best way to become familiar with them is to practice using them.

example

Solve the following limit: x→2lim⁡(2x3+x−6) ^{\lim}_{x \to 2} (2x^{3} + x - 6)

answer

Using the properties we learned above, this is quite easy:

x→2lim⁡(2x3+x−6) ^{\lim}_{x \to 2} (2x^{3} + x - 6)

= x→2lim⁡(2x3)+x→2lim⁡(x)−x→2lim⁡(6) = \ ^{\lim}_{x \to 2} (2x^{3}) + ^{\lim}_{x \to 2} (x) - ^{\lim}_{x \to 2} (6) using rule 3

=2[x→2lim⁡(x)]3+x→2lim⁡(x)−x→2lim⁡(6) = 2[ ^{\lim}_{x \to 2} (x)]^{3} + ^{\lim}_{x \to 2} (x) - ^{\lim}_{x \to 2} (6) using rules 4 and 7

=2[2]3+2−(6) = 2[2]^{3} + 2 - (6) using rules 1 and 2

=16+2−6 = 16 + 2 - 6 by evaluating

=12 = 12


∴ x→2lim⁡(2x3+x−6)=12\therefore \ ^{\lim}_{x \to 2} (2x^{3} + x - 6) = 12

Indeterminate limits

Some limits may not be directly solvable, as they may have a value of 00\frac{0}{0}. Limits with this type of value are indeterminate. 00\frac{0}{0} is not a valid answer, and you will need to use some clever strategies to solve them.

example: Solving a limit using factoring

Solve the following limit: x→3lim⁡x2−2x−3x−3 ^{\lim}_{x \to 3} \frac {x^2 - 2x - 3}{x - 3}

Evaluating using our current understanding of limits, you end up with 00\frac {0} {0} (try it yourself!), which is indeterminate.

answer

To evaluate the limit, you need to somehow make it so that the denominator is not equal to 00.


Luckily, we can factor out the (x−3) (x - 3) from the numerator and denominator:

x→3lim⁡x2−2x−3x−3 ^{\lim}_{x \to 3} \frac {x^2 - 2x - 3}{x - 3}

= x→3lim⁡(x+1)(x−3)x−3 = \ ^{\lim}_{x \to 3} \frac {(x + 1)(x - 3)}{x - 3}

= x→3lim⁡(x+1)(x−3)x−3 = \ ^{\lim}_{x \to 3} \frac {(x + 1) \cancel{(x - 3)} }{ \cancel{x - 3} }

= x→3lim⁡(x+1) = \ ^{\lim}_{x \to 3} (x + 1)

=(3)+1 = (3) + 1

=4 = 4


∴ x→3lim⁡x2−2x−3x−3=4 \therefore \ ^{\lim}_{x \to 3} \frac {x^2 - 2x - 3}{x - 3} = 4

example: Solving a limit by rationalizing

Solve the following limit: x→0lim⁡x+1−1x _{x \to 0}^{\lim} \frac {\sqrt{x+1}-1}{x}

Again, using what we learned earlier, this is not possible since we end up getting the indeterminate value 00\frac{0}{0} .

answer

Another powerful method to solve limits is to rationalize either the numerator or denominator.

x→0lim⁡x+1−1x _{x \to 0}^{\lim} \frac {\sqrt{x+1}-1}{x}

= x→0lim⁡x+1−1x×x+1+1x+1+1 = \ _{x \to 0}^{\lim} \frac {\sqrt{x+1}-1}{x} \times \frac{\sqrt{x+1}+1}{\sqrt{x+1}+1}

= x→0lim⁡x+1−1x(x+1+1) = \ _{x \to 0}^{\lim} \frac {x+1-1}{x(\sqrt{x+1}+1)}

= x→0lim⁡xx(x+1+1) = \ _{x \to 0}^{\lim} \frac {x}{x(\sqrt{x+1}+1)}

= x→0lim⁡xx(x+1+1) = \ _{x \to 0}^{\lim} \frac {\cancel{x}}{\cancel{x}(\sqrt{x+1}+1)}

= x→0lim⁡1x+1+1 = \ _{x \to 0}^{\lim} \frac {1}{\sqrt{x+1}+1}

=10+1+1 = \frac {1}{\sqrt{0+1}+1}

=12 = \frac {1}{2}


∴ x→0lim⁡x+1−1x=12 \therefore \ _{x \to 0}^{\lim} \frac {\sqrt{x+1}-1}{x} = \frac {1}{2}

example: Solving a limit by substitution

Solve the following limit: x→64lim⁡x3−4x−8 ^{\lim}_{x \to 64} \frac{\sqrt[3]{x} - 4}{\sqrt{x} - 8}

Yet again, trying to solve this limit results in an indeterminate 00\frac{0}{0} value.

answer

You can substitute certain values for another variable in order to rewrite the equation.


Take u=x16u = x^{\frac{1}{6}}, since it is a factor of both x3\sqrt[3]{x} and x\sqrt{x}.


First, calculate the value that uu should approach for the limit:

u=x16=(64)16=2 u = x^{\frac{1}{6}} = (64)^{\frac{1}{6}} = 2


Now, substitute all the values into the limit and solve it:

x→64lim⁡x3−4x−8 ^{\lim}_{x \to 64} \frac{\sqrt[3]{x} - 4}{\sqrt{x} - 8}

= u→2lim⁡u2−4u3−8 = \ ^{\lim}_{u \to 2} \frac{u^2 - 4}{u^3 - 8}

= u→2lim⁡(u−2)(u+2)(u−2)(u2+2u+4) = \ ^{\lim}_{u \to 2} \frac{(u-2)(u+2)}{(u-2)(u^2+2u+4)} \quad by difference of square and difference of cubes

= u→2lim⁡(u−2)(u+2)(u−2)(u2+2u+4) = \ ^{\lim}_{u \to 2} \frac{ \cancel{(u-2)} (u+2)}{ \cancel{(u-2)} (u^2+2u+4)}

= u→2lim⁡u+2u2+2u+4 = \ ^{\lim}_{u \to 2} \frac{u+2}{u^2+2u+4}

=(2)+2(2)2+2(2)+4 = \frac{(2)+2}{(2)^2+2(2)+4}

=412 = \frac{4}{12}

=13 = \frac{1}{3}


∴ x→64lim⁡x3−4x−8=13 \therefore \ ^{\lim}_{x \to 64} \frac{\sqrt[3]{x} - 4}{\sqrt{x} - 8} = \frac{1}{3}

Infinite limits

Take a look at the following graph of the function f(x)=1x f(x) = \frac {1} {x}

No description

As you know, f(x)=1x f(x) = \frac {1} {x} has an asymptote at x=0 x = 0 , meaning the function is not defined there.


The limit for this function at x=0 x = 0 is not defined. Let's take a look at the left and right limits:

As x→0 x \to 0 from the left, it nears negative infinity (−∞-\infty). so, x→0−lim⁡(1x)=−∞ ^{\lim}_{x \to 0^{-}} (\frac {1} {x}) = -\infty

As x→0 x \to 0 from the right, it nears positive infinity (+∞+\infty). so, x→0+lim⁡(1x)=+∞ ^{\lim}_{x \to 0^{+}} (\frac {1} {x}) = +\infty


Let's also take a look at the limit of this function as x→±∞ x \to \pm \infty :

As x→−∞ x \to - \infty , f(x)→0 f(x) \to 0 from the bottom

As x→+∞ x \to + \infty , f(x)→0 f(x) \to 0 from the top


Since both +∞+\infty and −∞-\infty are not real numbers, the left and right limits of f(x)f(x) are not defined, which also makes the limit of f(x)f(x) as xx approach 00 undefined as well. However, this is useful for graphing the function near these asymptotes, since we can determine how they behave as xx gets closer to them. The same applies for horizontal asymptotes as well.

example: Solving infinite limits

When solving an infinite limit for certain functions, you may end up with an indeterminate ±∞∞ \pm \frac {\infty} {\infty} answer.

Take x→∞lim⁡2(x2−9)x2−4 ^{\lim}_{x \to \infty} \frac {2(x^2 - 9)} {x^2 - 4} .

Using current methods, it is not possible to solve this limit.

answer

A handy trick to solving these limits is to divide the numerator and denominator by xx of the highest power.

If we expand the function f(x)=2(x2−9)x2−4 f(x) = \frac {2(x^2 - 9)} {x^2 - 4} , we get f(x)=2x2−18x2−4 f(x) = \frac {2x^2 - 18} {x^2 - 4} .


Now, divide the numerator by x2 x^2 :

f(x)=2x2−18x2−4×1x21x2=2−18x21−4x2 f(x) = \frac {2x^2 - 18} {x^2 - 4} \times \frac {\frac {1} {x^2}} {\frac {1} {x^2}} = \frac {2 - \frac {18} {x^2} } {1 - \frac {4} {x^2} }


You can easily solve the limit of this function by plugging in x=∞ x = \infty :

x→∞lim⁡2(x2−9)x2−4= x→∞lim⁡2−18x21−4x2 ^{\lim}_{x \to \infty} \frac {2(x^2 - 9)} {x^2 - 4} = \ ^{\lim}_{x \to \infty} \frac {2 - \frac {18} {x^2} } {1 - \frac {4} {x^2} }

    =2−18(∞)21−4(∞)2 \quad \quad \quad \quad \ \ \ \, = \frac {2 - \frac {18} {(\infty)^2} } {1 - \frac {4} {(\infty)^2} }

    =2−01−0 \quad \quad \quad \quad \ \ \ \, = \frac {2 - 0} {1 - 0} \quad because a∞=0 \frac {a} {\infty} = 0 for any constant aa

    =2 \quad \quad \quad \quad \ \ \ \, = 2


∴ x→∞lim⁡2(x2−9)x2−4=2 \therefore \ ^{\lim}_{x \to \infty} \frac {2(x^2 - 9)} {x^2 - 4} = 2