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Francesco Giannino » 5.Limit of functions


Limit laws

Assume that \lim_{x\rightarrow a}f(x) \,\,\text{and}\,\,\lim_{x\rightarrow a}g(x) exist, then

\lim_{x\rightarrow a}(f(x)\pm g(x))=\lim_{x\rightarrow a}f(x)\pm\lim_{x\rightarrow a}g(x)

\lim_{x\rightarrow a}(f(x)g(x))=\lim_{x\rightarrow a}f(x)\cdot \lim_{x\rightarrow a}g(x)

\lim_{x\rightarrow a}\frac{f(x)}{g(x)}= \frac{\lim_{x\rightarrow a}f(x)}{\lim_{x\rightarrow a}g(x)} \,\,\,\text{if}\,\,\lim_{x\rightarrow a}g(x)\neq 0

\lim_{x\rightarrow a}[f(x)]^{n}=[\lim_{x\rightarrow a}f(x)]^{n}

\lim_{x\rightarrow a}\sqrt[n]{f(x)}=\sqrt[n]{\lim_{x\rightarrow a}f(x)}

If n is even, we must additionally assume that \lim_{x\rightarrow a}f(x)> 0

Limit laws

In particolar, if f(x) is an algebraic function and f(a) is defined, then

\lim_{x\rightarrow a}f(x)=f(a)

So, in some cases it is possible to compute limits by substituting a for x.
Exercise 1. Compute the following limit \lim_{x\rightarrow 3}(x^{2}+2x+4).

Solution. By applying the limit laws repeatedly:

\lim_{x\rightarrow 3}(x^{2}+2x+4)=\lim_{x\rightarrow 3}(x^{2})+\lim_{x\rightarrow 3}(2x)+\lim_{x\rightarrow 3}(4)=

=(\lim_{x\rightarrow 3}x))^{2}+2\lim_{x\rightarrow 3}(x) +4 = 3^{2}+2\cdot 3 +4 = 19

Exercises

Exercise 2. Compute the following limit \lim_{x\rightarrow 3}\frac{\sqrt{x}+3x}{x^{2}-4x+4}
Solution. Given function is an algebraic function; thus:

\lim_{x\rightarrow 3}\frac{\sqrt{x}+3x}{x^{2}-4x+4}=  \frac{\sqrt{3}+3(3)}{3^{2}-4(3)+4}=<br />
\frac{\sqrt{3}+9}{1}=\sqrt{3}+9

Exercises

Finding limits by algebraic manipulation: \lim_{x\rightarrow 1}\frac{x^{2}-1}{x-1}

Exercise 3. Compute the following limit
Solution. Given function is an algebraic function and substituting x = 1, we see that both numerator and denominator go to zero (x = 1 is a root). Notice that in this case it is possible to cancel the fraction by an algebraic manipulation

\lim_{x\rightarrow 1}\frac{x^{2}-1}{x-1} =\lim_{x\rightarrow 1}\frac{(x-1)(x+1)}{x-1} =\lim_{x\rightarrow 1}(x+1)=2

Exercises

Exercise 4. Compute the following limit \lim_{x\rightarrow 3}\frac{x^{2}+x-12}{x^{2}-9}
Solution. Given function is an algebraic function and substituting x = 3, we notice that 3 is a root of numerator and denominator.
Factoring the numerator and the denominator, we get:

\lim_{x\rightarrow 3}\frac{x^{2}+x-12}{x^{2}-9}= \lim_{x\rightarrow 3}\frac{(x-3)(x+4)}{(x-3)(x+3)}= \lim_{x\rightarrow 3}\frac{(x+4)}{(x+3)}=\frac{7}{6}

Infinite limits

We recall that the notation \lim_{x\rightarrow a}f(x)= + \infty
means that the values of f(x) can be arbitrarily large by taking x sufficiently close to α but not equal to α.
We recall that the notation \lim_{x\rightarrow a}f(x)= - \infty
means that the values of f(x) can be arbitrarily large negative by taking x sufficiently close to α but not equal to α.
Examples.

\lim_{x\rightarrow 0^{+}}\frac{1}{x}= + \infty

\lim_{x\rightarrow 0^{-}}\frac{1}{x}= - \infty

\lim_{x\rightarrow 0}\frac{1}{x^{2}}=  \infty

Exercises

Exercise 5. Let the function \lim_{x\rightarrow 3}\frac{x^{2}+ 2}{x^{2}-3x+2}. and compute \lim_{x\rightarrow a^{-}}f(x) \,\,\text{and} \,\,\lim_{x\rightarrow a^{-}}f(x)
for each α at which f is not continuous.
Solution. Factoring the denominator, we get the following factors (x-1)(x-2), where in 1 and 2 the assigned function f is not continuous. We can record the signs of the factors as follows.


Exercises

Then

\lim_{x\rightarrow 1^{-}}f(x)= + \infty ,\lim_{x\rightarrow 1^{+}}f(x)= - \infty

\lim_{x\rightarrow 2^{-}}f(x)= - \infty ,\lim_{x\rightarrow 2^{+}}f(x)= + \infty

For example, as x → 1- , the numerator approaches 2, and the denominator approaches 0 while remaining positive. So the limit is + ∞.

Infinite limits

We recall that

\,\,\, \bullet \,\,\, \text{if}\,\,\lim_{x\rightarrow a}f(x)= + \infty \,\,\text{and}\,\,\lim_{x\rightarrow a}g(x)= + \infty \,\, \text{then}

\lim_{x\rightarrow a}(f(x)+g(x))=+ \infty

\,\,\, \bullet \,\,\, \text{if}\,\,\lim_{x\rightarrow a}f(x)= - \infty \,\,\text{and}\,\,\lim_{x\rightarrow a}g(x)= - \infty \,\, \text{then}

\lim_{x\rightarrow a}(f(x)+g(x))= - \infty<br />

\,\,\, \bullet \,\,\, \text{if}\,\,\lim_{x\rightarrow a}f(x)= L \,\,\text{and}\,\,\lim_{x\rightarrow a}g(x)= \pm \infty \,\, \text{then}

\lim_{x\rightarrow a}(f(x)+g(x))= \pm \infty

Infinite limits

We recall that:

  • The product of a finite limit and an infinite limit is infinite if the finite limit is not 0. That is

\text{if}\,\,\lim_{x\rightarrow a}f(x)= L\neq 0 \,\,\text{and}\,\,\lim_{x\rightarrow a}g(x)= + \infty \,\, \text{then}

\lim_{x\rightarrow a}f(x)g(x)=\begin{cases}     +\infty \,\,\text{if}\,\, L>0\\     -\infty \,\,\text{if}\,\, L<0\\    \end{cases}

\text{if}\,\,\lim_{x\rightarrow a}f(x)= L\neq 0 \,\,\text{and}\,\,\lim_{x\rightarrow a}g(x)= - \infty \,\, \text{then}

\lim_{x\rightarrow a}f(x)g(x)=\begin{cases}     -\infty \,\,\text{if}\,\, L>0\\     +\infty \,\,\text{if}\,\, L<0\\    \end{cases}

Infinite limits

We recall that:

  • The product of two infinite linits is infinite. That is:

\text{if}\,\,\lim_{x\rightarrow a}f(x)= + \infty \,\,\text{and}\,\,\lim_{x\rightarrow a}g(x)= + \infty \,\, \text{then}

\lim_{x\rightarrow a}f(x)g(x)= + \infty

\text{if}\,\,\lim_{x\rightarrow a}f(x)= + \infty \,\,\text{and}\,\,\lim_{x\rightarrow a}g(x)= - \infty \,\, \text{then}

\lim_{x\rightarrow a}f(x)g(x)= - \infty

\text{if}\,\,\lim_{x\rightarrow a}f(x)= - \infty \,\,\text{and}\,\,\lim_{x\rightarrow a}g(x)= - \infty \,\, \text{then}

\lim_{x\rightarrow a}f(x)g(x)= + \infty

Infinite limits

We recall that:

  • The quotient of a finite limit by an infinite linit is zero. That is:

\text{if}\,\,\lim_{x\rightarrow a}f(x)= L \neq 0 \,\,\text{and}\,\,\lim_{x\rightarrow a}g(x)= \pm \infty \,\, \text{then}

\lim_{x\rightarrow a}\frac{f(x)}{g(x)}= 0

Indeterminate limits

We recall that:

  • Limits of the form:

0\cdot \infty, \,\, \frac{\infty}{\infty}, \,\, \frac{0}{0}, \,\, \infty - \infty

are indeterminate. There is no rule for evaluating such a form; the limit must be examined more closely.

  • Limits of the form L/0 are also indeterminate and these limits for example can be evaluated as follows:

\lim_{x\rightarrow 0^{+}}\frac{1}{x}= + \infty \,\,\,\, \lim_{x\rightarrow 0^{-}}\frac{1}{x}= - \infty   \,\,\,\, \lim_{x\rightarrow 0}\frac{1}{x^{2}}=  \infty

Limits at infinity

We recall that:

  • If n is a positive integer number. Then

\lim_{x\rightarrow + \infty}\frac{1}{x^{n}}= 0

\lim_{x\rightarrow - \infty}\frac{1}{x^{n}}= 0

Exercises

Exercise 6. Compute the limit  \lim_{x\rightarrow + \infty}\frac{2x^{3}+3x+1}{4x^{3}+5x^{2}+7} if it exists.
Solution. Factoring out the largest power of x from the numerator and denominator, we get:

\lim_{x\rightarrow + \infty}\frac{2x^{3}+3x+1}{4x^{3}+5x^{2}+7}= \lim_{x\rightarrow + \infty}\frac{x^{3}(2+\frac{3}{x^{2}}+\frac{1}{x^{3}})} {x^{3}(4+\frac{5}{x}+\frac{7}{x^{3}})}=

=\lim_{x\rightarrow + \infty}\frac{2+\frac{3}{x^{2}}+\frac{1}{x^{3}}} {4+\frac{5}{x}+\frac{7}{x^{3}}}=\frac{2+0+0}{4+0+0}=\frac{1}{2}

Observation. When find limits of algebraic expressions at infinity, look at the highest degree terms.

Exercises

Exercise 7. Compute the limit \lim_{x\rightarrow + \infty}\frac{\sqrt{3x^{4}+5}}{x^{2}+7} if it exists.
Solution. As we noticed in the observation, when we have to find limits of algebraic expressions, we can look only at the highest degree terms. In this case, the numerator can be considered as follows:

\sqrt{3x^{4}+5} \sim \sqrt{3x^{4}} \sim \sqrt{3}x^{2}

Then, we expect that the limit of assigned function is \sqrt{3}.
In any case we can solve the assigned limit factoring out the largest power of x from the numerator and factoring out the largest power of x from the denominator under the root.

Exercises

\lim_{x\rightarrow + \infty}\frac{\sqrt{3x^{4}+5}}{x^{2}+7}= \lim_{x\rightarrow + \infty}\frac{\sqrt{x^{4}(3+\frac{5}{x^{4}})}}{x^{2}(1+\frac{7}{x^{2}}}=

=\lim_{x\rightarrow + \infty}\frac{x^{2}\sqrt{3+\frac{5}{x^{4}}}}{x^{2}(1+\frac{7}{x^{2}})}= \lim_{x\rightarrow + \infty}\frac{\sqrt{3+\frac{5}{x^{4}}}}{1+\frac{7}{x^{2}}}=

= \frac{\sqrt{3 + 0}}{1 + 0}= \sqrt{3}

Exercises

Exercise 8. Find the limit \lim_{x\rightarrow + \infty}\frac{5x^{3}+2x+1}{2x^{2}+3x+5} if it exists.
Solution. Looking directly only at the highest degree terms, we get:

\lim_{x\rightarrow + \infty}\frac{5x^{3}+2x+1}{2x^{2}+3x+5}= \lim_{x\rightarrow + \infty}\frac{5x^{3}}{2x^{2}}=

=\lim_{x\rightarrow + \infty}\frac{5x}{2}= + \infty

Exercises

Exercise 9. Find the limit  \lim_{x\rightarrow + \infty}\frac{4x^{2}+x+2}{2x^{4}+5x^{2}+3} if it exists.
Solution. Looking directly only at the highest degree terms, we get:

\lim_{x\rightarrow + \infty}\frac{4x^{2}+x+2}{2x^{4}+5x^{2}+3}= \lim_{x\rightarrow + \infty}\frac{4x^{2}}{2x^{4}}=

= \lim_{x\rightarrow + \infty}\frac{2}{x^{2}}= 0

Exercises

Exercise 10. Find the limit \lim_{x\rightarrow + \infty} (\sqrt{4x^{2}+ 15}-2x) if it exists.
Solution. This limit is of the form ∞ – ∞ which indeterminate and there is no rule for evaluating such a form; so we rationalize the numerator to get an expression that we can use the limit laws on:

\lim_{x\rightarrow + \infty} (\sqrt{4x^{2}+ 15}-2x)=   \lim_{x\rightarrow + \infty} (\sqrt{4x^{2}+ 15}-2x)  \cdot \frac{\sqrt{4x^{2}+ 15}+2x}{\sqrt{4x^{2}+ 15}+2x}=

=\lim_{x\rightarrow + \infty} \frac{4x^{2}+ 15-4x^{2}}{\sqrt{4x^{2}+ 15}+2x}=  =\lim_{x\rightarrow + \infty} \frac{15}{\sqrt{(4x^{2}+ 15)}+2x}= 0

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Progetto "Campus Virtuale" dell'Università degli Studi di Napoli Federico II, realizzato con il cofinanziamento dell'Unione europea. Asse V - Società dell'informazione - Obiettivo Operativo 5.1 e-Government ed e-Inclusion