Post on 15-Feb-2016
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Math 1241, Spring 2014Section 3.1, Part Two
Infinite Limits, Limits “at Infinity”Algebraic Rules for Limits
Infinite Limits
• If x is close to zero, then the function is close to what number? Here is the graph:
Infinite Limits
• IMPORTANT: DOES NOT EXIST!!!• There is a reason for this. As x approaches 0,
the function value keeps getting larger, and never approaches any particular value.
• Notation: • But you CANNOT treat the infinity symbol as
though it were an ordinary number.
Examples of Infinite Limits
Convince yourself (possibly by drawing a graph) that the following are true:
For the left-sided limit, the means that the function value continues to decrease, and does not approach any particular value.
𝑓 (𝑥 )=1𝑥
Limits “at Infinity”
• On the previous graph, what happens to the value of f(x) as x gets “larger and larger?”– On the graph: Further and further to the right.
• Similar question: what happens to the value of f(x) as x gets “more and more negative?”– On the graph: Further and further to the left.
• In previous courses, these questions were related to horizontal asymptotes of the graph.
Limits “at Infinity”
The notation:
means, “As the value of x gets larger and larger, the value of f(x) approaches the number L.”
In similar fashion:
means, “As the value of x gets more and more negative, the value of f(x) approaches the number L.”
Example:
Infinite Limits “at Infinity”
• We can also have infinite limits “at infinity.” For example:
• This means, “As the value of x gets larger and
larger, the value of becomes more and more negative.” See the graph on the next slide.
• NOTE: If a limit “equals” + or , that limit DOES NOT EXIST. The notation allows us to indicate why the limit does not exist.
𝑓 (𝑥 )=1−𝑥2
Algebraic Rules for Limits
• For most “ordinary” algebraic functions, you can “plug in x = a” to evaluate .
• In particular, this works for:– Polynomials (example: )– Rational functions, except when the denominator
is zero: (example: for )– Exponential functions (example: )– Logarithms (example: )
Simple Algebraic Examples
Evaluate .• Solution: Since is a polynomial, you can
evaluate the limit by plugging in x = 3.• .• We can confirm this with a graph, see the next
slide.
lim𝑥→ 3
(𝑥3−3 𝑥2+2𝑥+1 )=7
Exercises
Evaluate the following limits, and compare your results with the previous graph.•
These should be very easy exercises!
Exercise: Rational Functions
Let Evaluate the limits:
The first limit should be very easy. The second requires more work.
Solutions
• Since is a rational function (ratio of two polynomials), we can evaluate by plugging in x = 1:
• What happens when you plug in x = 2 ?
𝑓 (𝑥 )= 𝑥2−2𝑥𝑥2+𝑥−6
lim𝑥→ 2
𝑥2−2𝑥𝑥2+𝑥−6
=? ?
• Many graphing programs do not detect the “hole in the graph” when x = 2.
• When our function has a zero denominator, we can try to factor numerator/denominator, and hope that the zero factor cancels.
• HINT: In this case, the numerator and denominator are zero at x = 2, so there should be a factor of (x-2).
lim𝑥→ 2
𝑥2−2𝑥𝑥2+𝑥−6
=? ?
• When evaluating the limit as , we may assume . This allows us to cancel the zero factor from the denominator.
• If you graph , you will see that it has the same graph as the function above (with the hole filled in at ).
• We can now plug in x = 2, giving a limit of .
An Important Result
• For this type of scenario, we have the following:If whenever , then
• In other words, the value of (even if it is undefined) does not have any effect on the value of .
• Note: This is Rule #7, pg. 128. We saw a graphical version of this last time.
For more complicated functions, we can often evaluate limits with the following rules (pg. 128)
Roots/Fractional Exponents
• We can usually “take the limit symbol through the radical sign,” but we must be careful with even roots (including square roots).
• If n is an odd positive integer, then
provided that exists.• If n is even, the above rule works when .
Example:
The idea is to use the limit rules to break this down into limits of more simple functions.
(provided that both limits on the right exist)• (why?)
(provided the limit under the radical exists and is positive)• (why?)
lim𝑥→2
(3¿¿ 𝑥+√𝑥2+5¿)=9+3=12¿¿