Evaluating Limits Analytically (College Board AP® Calculus AB)
Study Guide
Written by: Roger B
Reviewed by: Dan Finlay
Substituting to find limits
How can I use substitution to find limits?
If a function is continuous on an open interval, and if is a number contained in that interval, then
I.e. for a continuous function, the limit at a point is equal to the value of the function at that point
This idea can be extended to a point that is on or 'just outside' the border of an interval on which a function (or part of a function) is defined
This is especially useful for one-sided limits at the 'joins' of piecewise functions
E.g. the function defined by
Note that and are both continuous functions on all the real numbers
is on the border for the piece of , and 'just outside' the border for the piece of
So by substituting,
and
Worked Example
Let be the function defined by
Find:
(a)
Answer:
belongs to the piece of the function, and is continuous on the open interval
So the limit is just equal to the value of the function at
(b)
For this limit to exist, the limits from the left and right must both exist and be equal
is 'just outside' the border for the piece of , and on the border for the piece of
So use substitution to find the one-sided limits
Those limits are equal, so that is the value of the (two-sided) limit at
Simplifying to find limits
How can I simplify functions to find limits?
Sometimes trying to evaluate a limit by substitution gives an answer that is not defined
E.g. returning a value like
In these cases, sometimes the equation of the function can be simplified
For example by factorising and cancelling common factors
After this substitution may give a well-defined value for the limit
E.g. let be the function defined by
This function is undefined at , because
But to find the limit at we don't care about the function's value at
We only care about it's behavior near
So we can cancel the common factor and proceed like this:
The 'vertical bar notation' means to evaluate the expression to the left of the bar at
Note that the limit can be well-defined even at a point where the function itself is undefined!
Also note that and are not the same function
The function is defined at , and the function isn't
But for purposes of calculating limits, canceling common factors to make 'look like' is perfectly okay
Not all limits for which substitution gives can be resolved this way
See the L'Hospital's Rule study guide for another method to deal with such limits
Worked Example
Let be the function defined by .
Find .
Answer:
Attempting to evaluate this limit by substitution doesn't work:
Instead, factorise the numerator and denominator, simplify the fraction, and use substitution to evaluate the limit in the simplified form
Multiplying by conjugates to find limits
How can I use multiplying by conjugates to help find limits?
A quotient involving surds (i.e. square roots) can sometimes be rewritten by multiplying it top and bottom by a conjugate of either the numerator or denominator
The conjugate of is
and the conjugate of is
For example,
This can be used to find limits where just using substitution gives an undefined expression like
Worked Example
Let be the function defined by .
Find .
Answer:
Attempting to evaluate this limit by substitution doesn't work:
Instead, multiply the top and bottom by the conjugate of the denominator, then simplify the fraction
Now use substitution to evaluate the limit in the simplified form
Multiplying by reciprocals to find limits
How can I use multiplying by reciprocals to find a limit at infinity?
A quotient involving powers of x can be rewritten by multiplying its top and bottom by the reciprocal of the highest power of x in the numerator or denominator
For example,
This can be used to find a limit at infinity
because if is any constant and is a positive integer, then
Worked Example
Let be the function defined by .
Find .
Answer:
The numerator and denominator of the fraction both tend to infinity as x tends to infinity, but is not defined as a limit (see the study guide on L'Hospital's Rule for another way to deal with limits like this)
Instead, multiply the top and bottom by the reciprocal of the highest power,
Now take the limit of that rewritten form
All the terms in the numerator and denominator tend towards zero as x tends towards infinity, except for the constant terms (4 in the numerator and 5 in the denominator)
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