- WELCOME TO A FORMAL DEFINITION OF LIMITS.
THE GOALS OF THIS VIDEO ARE TO GIVE A FORMAL DEFINITION
OF A LIMIT AND ALSO TO PROVE A SPECIFIC LIMIT EXISTS.
LET'S DO A QUICK REVIEW OF THE IDEA OF A LIMIT INFORMALLY.
THE LIMIT AS X APPROACHES 3 OF F OF X EXISTS
IF AS WE APPROACH 3, IDENTIFIED BY THIS VERTICAL LINE,
FROM BOTH THE LEFT SIDE AND THE RIGHT SIDE,
AND WE APPROACH THE SAME FUNCTION VALUE,
THE LIMIT EXISTS AND THE LIMIT IS EQUAL
TO THAT FUNCTION VALUE THAT WE'RE APPROACHING.
SO, FROM THIS GRAPH, WE CAN SEE AS WE APPROACH 3
FROM THE LEFT, WE'RE APPROACHING A FUNCTION VALUE
OR A Y VALUE OF 2.
AND THE SAME IS TRUE AS WE APPROACH FROM THE RIGHT.
THEREFORE, THIS LIMIT IS EQUAL TO 2 EVEN THOUGH THIS FUNCTION
DOES NOT EXIST THAT X = 3, AS WE CAN SEE BY THIS HOLE.
NOW LET'S TAKE A LOOK AT THE FORMAL DEFINITION OF A LIMIT.
IF WE LET F OF X BE A DEFINED FUNCTION
ON AN OPEN INTERVAL CONTAINING C,
AND IT DOESN'T NEED TO BE DEFINED AT C
AS WE SAW FROM THE PREVIOUS GRAPH,
THE LIMIT AS X APPROACHES C OF F OF X = L MEANS THAT
FOR EVERY EPSILON GREATER THAN ZERO, THERE EXISTS
A DELTA GREATER THAN ZERO SUCH THAT,
NOW, REMEMBER, ABSOLUTE VALUE MEANS DISTANCE,
SO THE DISTANCE BETWEEN X AND C IS GREATER THAN ZERO
AND LESS THAN DELTA.
SO, DELTA IS A HORIZONTAL DISTANCE.
THEN DISTANCE BETWEEN THE FUNCTION
AND THE LIMIT IS LESS THAN EPSILON,
SO, EPSILON WILL BE A VERTICAL DISTANCE FROM L.
SO THIS MEANS IN ORDER TO PROVE A LIMIT EXISTS,
WE NEED A RELATIONSHIP BETWEEN DELTA AND EPSILON.
SO, WHAT THIS IS SAYING IS THAT FOR ANY VERTICAL DISTANCE
EPSILON FROM L THERE NEEDS TO BE
A HORIZONTAL DISTANCE DELTA SO THAT THE DISTANCE
BETWEEN X AND C IS LESS THAN DELTA SUCH THAT
THE DISTANCE BETWEEN F OF X AND L IS LESS THAN EPSILON.
TO DEVELOP THIS IN A PROBLEM WE'RE ACTUALLY GOING
TO WORK BACKWARDS, BUT LET'S TAKE A LOOK
AT THIS GRAPHICALLY FIRST.
SO IF WE GENERALIZE THAT FIRST GRAPH THAT WE LOOKED AT,
LET 3 ON THE X-AXIS = C AND LET 2 ON THE Y-AXIS = L,
AND THEN LET EPSILON BE THIS VERTICAL DISTANCE HERE,
THIS INTERVAL WOULD BE L - EPSILON
ALL THE WAY TO L + EPSILON.
AND IF WE LET DELTA BE THIS VERTICAL DISTANCE
HERE AND HERE, THIS VERTICAL INTERVAL
WOULD BE C - DELTA ALL THE WAY TO C + DELTA.
AND THE IDEA OF THE FORMAL DEFINITION
IS THAT IF WE'RE GIVEN SOME EPSILON THAT PRODUCES
A VERTICAL INTERVAL AROUND OUR LIMIT L,
THERE MUST BE SOME HORIZONTAL DISTANCE DELTA
SUCH THAT THE INPUTS INTO THE FUNCTION
IN THIS INTERVAL WOULD HAVE TO PRODUCE OUTPUTS
IN THE INTERVAL FROM L PLUS OR MINUS EPSILON.
SO THE KEY TO WRITING THESE PROOFS
IS DEVELOPING A RELATIONSHIP BETWEEN EPSILON AND DELTA.
SO, LET'S TAKE A LOOK AT OUR FIRST EXAMPLE.
WE HAVE THE LIMIT AS X APPROACHES 6
OF 1/3X + 1 = 3.
NOW, OUR FUNCTION IS LINEAR, SO WE CAN PERFORM
BASIC SUBSTITUTION TO VERIFY THIS LIMIT IS TRUE.
BUT WE ACTUALLY WANT TO PROVE IT,
SO WE NEED TO DEVELOP A RELATIONSHIP BETWEEN EPSILON
AND DELTA IN ORDER TO WRITE THE PROOF.
SO WHAT WE'RE GOING TO DO IS START WITH WHAT WE WANT
TO SHOW THAT THE ABSOLUTE VALUE OF F OF X - L
IS LESS THAN EPSILON.
AND THEN WE'RE GOING TO MANIPULATE
THIS ABSOLUTE VALUE TO MAKE IT MATCH
THE ABSOLUTE VALUE OF X - C.
AND BY DOING THAT, WE'LL DETERMINE THE RELATIONSHIP
BETWEEN EPSILON AND DELTA.
SO WE'LL START WITH THE ABSOLUTE VALUE
OF F OF X - L IS LESS THAN EPSILON.
AND NOW WE'LL SUBSTITUTE 1/3X + 1 FOR F OF X,
AND WE'LL SUBSTITUTE 3 FOR L.
SO WE'LL HAVE THE ABSOLUTE VALUE OF 1/3X + 1 - 3
LESS THAN EPSILON.
AND NOW WE WANT TO MANIPULATE THIS
SO THIS BECOMES THE ABSOLUTE VALUE OF X - C,
WHICH IN THIS CASE WOULD BE ABSOLUTE VALUE OF X - 6.
SO LET'S FIRST CLEAR THE PARENTHESES.
WE HAVE THE ABSOLUTE VALUE OF 1/3X + 1 - 3, THAT'S -2.
WE STILL HAVE EPSILON ON THE RIGHT.
NOW, IF WE WANT THIS TO BE X INSTEAD OF 1/3X,
WE COULD MULTIPLY BOTH SIDES OF THE INEQUALITY BY 3.
SO, NOW THAT WE HAVE 3 EPSILON, NOW LET'S GO AHEAD
AND DISTRIBUTE THIS 3 INTO THE ABSOLUTE VALUE.
SO 3 x 1/3X IS X AND 3 x 2 IS 6,
SO WE HAVE -6 LESS THAN 3 x EPSILON.
NOW, REMEMBER WE'RE TRYING TO SHOW A RELATIONSHIP
BETWEEN EPSILON AND DELTA, AND NOTICE HOW
THE ABSOLUTE VALUE OF X - C IS LESS THAN DELTA,
BUT WE'VE SHOWN THAT THE ABSOLUTE VALUE OF X - 6
IS LESS THAN 3 EPSILON.
SO THIS TELLS US THAT WE CAN LET 3 EPSILON = DELTA,
AND THEN FROM HERE WE CAN WRITE OUR PROOF.
SO TO DETERMINE THE RELATIONSHIP,
WE KIND OF WORKED OUR WAY FROM RIGHT TO LEFT.
NOW THAT WE FOUND THE RELATIONSHIP
BETWEEN EPSILON AND DELTA, WE'LL WRITE OUR PROOF
IN THE OTHER DIRECTION FROM LEFT TO RIGHT.
SO IN ORDER TO PROVE THIS LIMIT, WE'LL START WITH ZERO
LESS THAN THE ABSOLUTE VALUE OF X - C LESS THAN DELTA.
AND NOW WE'LL PERFORM THE SUBSTITUTION.
SO WE'LL HAVE THE ABSOLUTE VALUE OF X - C, BUT C IS 6,
LESS THAN DELTA.
AND NOW WE'RE GOING TO GO AHEAD AND REPLACE DELTA
WITH 3 EPSILON.
AND NOW WE'RE GOING TO MANIPULATE THIS TO END UP
WITH THE ABSOLUTE VALUE OF THE QUANTITY 1/3X + 1 - 3
IS LESS THAN EPSILON, WHICH WILL BE THE ABSOLUTE VALUE
OF F OF X - L LESS THAN EPSILON.
SO NOTICE OVER HERE WE MULTIPLIED BY 3,
SO NOW WHAT WE'RE GOING TO DO IS FACTOR 3 OUT OF X - 6.
WELL, IF WE FACTOR 3 OUT, THIS WOULD BE 1/3X,
IF WE FACTOR A 3 OUT OF 6, THAT'LL BE 2.
AND NOW WE'LL GO AHEAD AND PULL THE 3
OUT OF THE ABSOLUTE VALUE.
AND NOW WE'LL GO AHEAD AND DIVIDE EVERYTHING BY 3.
SO IF WE DIVIDE THIS BY 3, WE'RE GOING TO HAVE
THE ABSOLUTE VALUE OF 1/3X - 2
IS LESS THAN EPSILON.
SO YOU KIND OF SEE WHERE WE'RE HEADED.
AND NOW JUST TO MAKE IT FIT THE FORM OF F OF X - L,
WE KNOW F OF X IS 1/3X + 1, SO WE'LL GO AHEAD
AND REWRITE THIS AS 1/3X + 1,
BUT IT STILL HAS TO BE MINUS 2 SO WE'LL HAVE TO SUBTRACT 3,
IS LESS THAN EPSILON, AND NOW WE'RE DONE
BECAUSE WE'VE SHOWN THAT ZERO IS LESS THAN THE ABSOLUTE VALUE
OF THIS IS F OF X - OUR LIMIT 3
IS LESS THAN EPSILON.
SO, AGAIN, TO DEVELOP THIS RELATIONSHIP WE WORKED
OUR WAY FROM RIGHT TO LEFT, AND ONCE WE FOUND
THE RELATIONSHIP BETWEEN EPSILON AND DELTA,
WE WORKED OUR WAY FROM LEFT TO RIGHT
IN ORDER TO WRITE THE PROOF.
NOTICE HOW WE STARTED WITH THE ABSOLUTE VALUE OF X - C
IS GREATER THAN ZERO AND LESS THAN DELTA.
AND WE SHOWED THAT IF THAT WAS TRUE THEN THE ABSOLUTE VALUE
OF F OF X - L WAS LESS THAN EPSILON.
SO THERE'S THE PROOF THAT OUR LIMIT IS EQUAL TO 3.