- WELCOME TO DETERMINING LIMITS.
THE GOAL OF THIS VIDEO, DETERMINE THE LIMITS
OF FUNCTIONS IF THEY EXIST USING A VARIETY OF TECHNIQUES.
IN THE VIDEO ENTITLED "INTRODUCTION TO LIMITS,"
WE DISCUSSED WHAT A LIMIT IS AND HOW TO DETERMINE A LIMIT
GRAPHICALLY AND NUMERICALLY.
THIS VIDEO WILL FOCUS ON USING ANALYTICAL METHODS.
THERE ARE MANY LIMIT PRINCIPLES THAT YOU SHOULD REVIEW
AND BE AWARE OF WHEN DETERMINING LIMITS.
IF YOU HAVEN'T, PLEASE REVIEW THEM.
YOU WILL FIND THERE ARE MANY LIMITS
THAT CAN BE FOUND BY DIRECT SUBSTITUTION.
LET'S TAKE A LOOK AT THE LIMIT AS X APPROACHES
4 OF X SQUARED - 6X + 9.
WE KNOW WE HAVE TO APPROACH THE VALUE OF 4 FROM BOTH SIDES,
THE LEFT AND THE RIGHT.
IF WE TAKE A LOOK AT THE GRAPH FIRST,
WE CAN SEE THAT AS WE APPROACH FROM THE RIGHT
AND WE APPROACH FROM THE LEFT, WE'RE ESSENTIALLY APPROACHING
THIS POINT ON THE FUNCTION.
SINCE THIS POINT IS ON THE FUNCTION,
WE CAN FIND A Y VALUE OF THAT FUNCTION BY PERFORMING
DIRECT SUBSTITUTION, SO ESSENTIALLY THIS LIMIT
WILL BE EQUAL TO F4.
IF WE CALCULATE F OF 4 WE'LL SEE WE'LL OBTAIN
THE VALUE OF +1, AS WE SEE ON THE GRAPH.
F OF 4 IS EQUAL TO 4 SQUARED - 6 x 4 + 9,
WHICH IS EQUAL TO 1, WHICH ALSO GIVES US
THE VALUE OF THIS LIMIT.
AGAIN, VISUALLY WE CAN SEE THE Y VALUE OF THIS POINT
IS EQUAL TO 1.
IN FACT, IF P OF X IS A POLYNOMIAL FUNCTION,
THE LIMIT OF P OF X AS X APPROACHES A
IS EQUAL TO P OF A.
NOW, LET'S TAKE A LOOK AT THIS LIMIT
AS X APPROACHES -3.
HERE IS THE LINE X = -3.
AS WE APPROACH FROM THE RIGHT AND THE LEFT,
WE CAN SEE WE'RE APPROACHING THIS HOLE IN THE FUNCTION.
WE CAN ALSO SEE THAT THIS Y VALUE WOULD BE -6,
SO WE CAN PREDICT THIS LIMIT WOULD EQUAL -6.
HOWEVER, OUR GOAL IS TO FIND AN ANALYTICAL METHOD
THAT WOULD MAKE IT EASIER TO FIND THIS LIMIT.
HOWEVER, THE PROBLEM IS FORMING DIRECT SUBSTITUTION
IN THIS FORM WOULD RESULT IN A ZERO IN THE DENOMINATOR.
HOWEVER, IF WE SIMPLIFY THIS FUNCTION FIRST,
X SQUARED - 9 IS A DIFFERENCE OF SQUARES,
WHAT WE'LL FIND IS THAT THERE'S A COMMON FACTOR OF X + 3
IN THE NUMERATOR AND DENOMINATOR.
IF WE SIMPLIFY THIS OUT, WE'RE LEFT WITH A LIMIT
AS X APPROACHES -3 OF X - 3, WHICH WE CAN NOW JUST PERFORM
DIRECT SUBSTITUTION,
WE'D HAVE -3 - 3 WHICH IS EQUAL TO -6.
THIS ALGEBRAIC METHOD IS AN ALTERNATIVE
TO ANALYZING THE GRAPH.
AND, IN FACT, SIMPLIFYING A FUNCTION
BEFORE FINDING A LIMIT IS OFTEN VERY HELPFUL.
LET'S TAKE A LOOK AT ANOTHER.
THE LIMIT AS X APPROACHES 4 OF THIS FUNCTION WOULD PRODUCE
A GRAPH AS WE SEE BELOW.
AS WE APPROACH X = 4 FROM THE LEFT AND FROM THE RIGHT,
AGAIN, WE SEE THAT WE WOULD BE APPROACHING
THIS HOLE IN THE FUNCTION, WHICH OBVIOUSLY HAS A Y VALUE
SOMETHING LESS THAN 1.
LET'S TAKE A LOOK IF WE CAN MANIPULATE THIS FUNCTION
ALGEBRAICALLY SO WE CAN PERFORM DIRECT SUBSTITUTION.
WE SEE IN ITS CURRENT FORM IF WE SUBBED IN 4
WE'D HAVE A DENOMINATOR OF 0.
ANOTHER TECHNIQUE THAT IS OFTEN USED
IS RATIONALIZING THE NUMERATOR, WHICH MEANS WE'RE GOING
TO MULTIPLY THE NUMERATOR BY ITS CONJUGATE
WHICH WOULD BE THE SQUARE ROOT OF X + 2.
LET'S SEE WHAT THIS DOES.
OKAY, IF WE MULTIPLY OUT THE NUMERATOR,
WE'RE GOING TO GET X - 4.
THE TRICK HERE IS THAT WE DO NOT WANT TO MULTIPLY
THE DENOMINATOR TOGETHER.
WE'RE GOING TO LEAVE IT IN FACTORED FORM,
AND WHAT WE'LL FIND IS THERE'S A COMMON FACTOR OF X - 4.
SO, IF WE SIMPLIFY THIS FUNCTION WE'RE GOING TO HAVE THE LIMIT
AS X APPROACHES 4 OF 1 OVER THE SQUARE ROOT OF X + 2.
NOW, WE CAN DO DIRECT SUBSTITUTION.
OF COURSE THE SQUARE ROOT OF 4 WOULD BE 2.
1/2 + 2 WOULD GIVE US 1/4, THE VALUE OF THIS LIMIT.
THIS TECHNIQUE IS CALLED RATIONALIZING THE NUMERATOR.
OKAY, NOW WE HAVE A SQUARE ROOT FUNCTION.
THIS FIRST LIMIT IS AS X APPROACHES 5 HERE,
AGAIN, WE CAN SEE THE FUNCTION IS NICE AND CONTINUOUS
IN THIS AREA.
THE POINT THAT IT'S APPROACHING IS ON THE FUNCTION,
SO WE CAN FIND THIS LIMIT BY PERFORMING DIRECT SUBSTITUTION.
THIS LIMIT WILL BE EQUAL TO THE SQUARE ROOT OF 5 + 4 - 2,
OBVIOUSLY THE SQUARE ROOT OF 9 WOULD BE 3, 3 - 2 IS EQUAL TO 1.
THIS LIMIT IS EQUAL TO 1, WHICH, AGAIN,
WE CAN VERIFY GRAPHICALLY.
OKAY. THIS NEXT LIMIT WE'RE APPROACHING -4.
WE'RE APPROACHING THIS YELLOW VERTICAL LINE.
HOWEVER, THIS IS A ONE-SIDED LIMIT.
WE'RE ONLY APPROACHING -4 FROM THE POSITIVE SIDE
OR THE RIGHT SIDE.
SO WOULD BE APPROACHING IT FROM THIS DIRECTION.
AND YOU CAN SEE AS WE GET CLOSER AND CLOSER
TO -4 WE ARE APPROACHING THIS POINT HERE,
WHICH IS A POINT ON THE FUNCTION, WHICH, AGAIN,
IS ANOTHER SIGN THAT WE CAN PERFORM DIRECT SUBSTITUTION.
LET'S TRY IT AND SEE WHAT WE GET.
SQUARE ROOT OF -4 + 4 - 2.
WELL, OF COURSE THE SQUARE ROOT OF 0 IS 0,
0 - 2 WOULD BE -2.
SO THE LIMIT AS X APPROACHES -4 FROM THE RIGHT IS EQUAL TO -2.
AGAIN, GRAPHICALLY THAT SEEMS TO BE CORRECT.
WE'RE APPROACHING THIS RED POINT HERE
AND THE Y VALUE IS -2.
ON THIS LAST ONE-SIDED LIMIT, WE'RE ALSO APPROACHING -4,
BUT FROM THE LEFT.
BY LOOKING AT THE GRAPH, THERE AREN'T ANY POINTS
AT THE LEFT OF -4.
THEREFORE, THIS LIMIT WOULD NOT EXIST.
ANOTHER REASON THIS LIMIT DOES NOT EXIST IS THAT THERE ARE
NO VALUES IN THE DOMAIN OF THIS FUNCTION
TO THE LEFT OF -4.
ALL RIGHT, LET'S TAKE A LOOK AT ONE MORE.
DETERMINE THE LIMITS OF THE GIVEN FUNCTION.
THIS FIRST LIMIT AS X APPROACHES 3 FROM THE LEFT,
SO WE'LL BE APPROACHING THIS, SO WE'LL BE APPROACHING
THIS YELLOW VERTICAL LINE, BUT IN THIS CASE
ONLY FROM THE LEFT.
SO AS I APPROACH FROM THE LEFT,
YOU CAN SEE THAT THE GRAPH GOES DOWN INDEFINITELY.
SO THE Y VALUES WOULD BE APPROACHING NEGATIVE INFINITY.
AS DISCUSSED IN A PREVIOUS VIDEO,
NEGATIVE INFINITY DOES NOT EXIST.
THE NEXT ONE-SIDED LIMIT WE'RE APPROACHING 3 FROM THE RIGHT.
SO IF WE APPROACH 3 FROM THE RIGHT, YOU CAN SEE
THE GRAPH GOES UP INDEFINITELY.
OUR Y VALUES ARE INCREASING WITHOUT BOUND.
THEREFORE, OUR LIMIT WOULD BE EQUAL TO POSITIVE INFINITY
IN THIS CASE.
AGAIN, THAT LIMIT DOES NOT EXIST.
I'D LIKE TO TAKE A MOMENT AND VERIFY
THESE TWO ONE-SIDED LIMITS IN THE TABLE FEATURE
OF THE GRAPHING CALCULATOR.
I'VE ALREADY TYPED IN THE FUNCTION INTO Y1.
NOW I'M GOING TO HIT SECOND GRAPH.
THESE FIRST FOUR ROWS OF X VALUES ARE VALUES
THAT APPROACH 3 FROM THE LEFT LIKE THIS FIRST ONE-SIDED LIMIT.
BY LOOKING AT THE Y VALUES WE CAN EASILY SEE THAT
THESE Y VALUES WOULD BE APPROACHING NEGATIVE INFINITY.
THESE LAST THREE ROWS APPROACH 3 FROM THE RIGHT
OR FROM VALUES GREATER THAN 3.
AND, AGAIN, BY LOOKING AT THE Y VALUES
WE CAN SEE THAT IT WOULD BE APPROACHING POSITIVE INFINITY.
THE LAST LIMIT IS THE LIMIT AS X APPROACHES 3,
WHICH MEANS WE HAVE TO APPROACH 3 FROM BOTH SIDES.
AND WE CAN SEE FROM THE TWO PREVIOUS
ONE-SIDED LIMITS THAT THOSE DON'T EXIST.
OF COURSE THIS ONE IS NOT GOING TO EXIST EITHER.
OKAY, I HOPE THAT HELPS EXPLAIN SEVERAL TECHNIQUES
YOU CAN USE TO DETERMINE LIMITS.
THANK YOU FOR WATCHING AND HAVE A GOOD DAY.