- WELCOME TO A VIDEO ON
THE LIMITS OF FUNCTIONS OF TWO VARIABLES.
LET'S START BY REVIEWING WHAT WE KNOW
ABOUT LIMITS OF FUNCTIONS IN ONE VARIABLE.
IF WE WANT TO CONSIDER THE LIMIT OF F OF X AS X APPROACHES C,
IT'S EQUAL TO L ONLY IF THE LIMIT
FROM THE LEFT AND RIGHT OF C ALSO EQUAL L.
SO IT WAS IMPORTANT THAT WE HAD TO CONSIDER THE LIMIT
FROM BOTH SIDES TO DETERMINE IF THE LIMIT EXISTS.
NOW, IF WE CONSIDER A LIMIT OF A FUNCTION IN TWO VARIABLES
AS XY APPROACHES AB,
THE LIMIT OF F OF XY = L
IF THE LIMIT FROM ALL PATHS APPROACHING AB EXISTS
AND ARE EQUAL TO L.
SO WHAT THIS IS SAYING IS
IF THIS IS THE POINT AB ON THE SURFACE,
IN ORDER FOR THAT LIMIT TO EXIST,
IT DOESN'T MATTER WHICH DIRECTION
WE APPROACH THIS POINT.
THE LIMIT MUST ALWAYS EQUAL L.
SO WE COULD APPROACH FROM THIS DIRECTION HERE.
WE COULD APPROACH FROM THIS DIRECTION HERE,
AND THEN THERE'S AN INFINITE NUMBER OF PATHS
THAT WE COULD CONSIDER TO DETERMINE IF THAT LIMIT EXISTS.
SO WE HAVE TO FIGURE OUT HOW ARE WE GOING TO DETERMINE A LIMIT
IF THERE'S POTENTIALLY AN INFINITE NUMBER
OF PATHS TO CONSIDER?
WELL, THERE IS ONE SHORTCUT IN SOME CASES.
REMEMBER, WITH FUNCTIONS OF ONE VARIABLE,
IF THE FUNCTION WAS CONTINUOUS AROUND THE VALUE OF C,
WE COULD DETERMINE THE LIMIT BY PERFORMING DIRECT SUBSTITUTION.
REMEMBER, WE TOOK THIS VALUE OF C
AND JUST PLUGGED IN INTO THE FUNCTION
IF THAT FUNCTION WAS CONTINUOUS AROUND THAT VALUE.
WELL, THE SAME IS TRUE FOR FUNCTIONS OF TWO VARIABLES.
IF THE FUNCTION IS CONTINUOUS,
THEN THE LIMIT AS XY APPROACHES AB OF F OF XY = F OF AB,
IF WE KNOW THE SURFACE IS NICE AND SMOOTH AND CONTINUOUS.
SO IF YOU HAVE SOME GRAPHING SOFTWARE
WHERE YOU CAN GRAPH IN 3D, THAT CAN OFTEN BE HELPFUL.
IT'S IMPORTANT TO REMEMBER
THAT IF A FUNCTION IS NOT CONTINUOUS AT AB,
THIS DOES NOT MEAN THE LIMIT DOES NOT EXIST.
WE WILL NEED TO CONSIDER APPROACHING AB
FROM DIFFERENT PATHS.
SO LET'S TALK ABOUT THE PATHS FOR A MOMENT.
IF XY IS APPROACHING THE POINT AB,
WE COULD CONSIDER LETTING X = A, Y = B.
WE MAY ALSO CONSIDER THE PATHS OF Y = X,
MAYBE EVEN Y = -X, Y = X SQUARED X = Y SQUARED AND SO ON
AND THESE MAY HELP US DETERMINE IF THE LIMIT EXISTS OR NOT.
LET'S GO AHEAD AND TAKE A LOOK AT SOME EXAMPLES.
HERE WE HAVE THE LIMIT AS XY APPROACHES (1, 1)
OF THIS FUNCTION IN TWO VARIABLES.
NOW, OVER HERE WE DO HAVE THE GRAPH OF THE SURFACE,
AND WE CAN QUICKLY SEE THERE IS SOME DISCONTINUITY.
LET'S SEE IF WE CAN LOCATE WHERE THE POINT (1, 1) WOULD BE.
REMEMBER, WE'D MOVE OUT ONE ALONG THE X AXIS,
OUT ONE AROUND THE Y AXIS, AND THEN UP TO THE SURFACE.
SO IT LOOKS LIKE THIS POINT WOULD BE
SOMEWHERE OVER HERE ON THE SURFACE,
AND YOU CAN SEE AROUND THE REGION OF THIS POINT
THE SURFACE IS NICE AND SMOOTH AND CONTINUOUS.
THIS IS AN INDICATION THAT WE CAN JUST GO AHEAD
AND PERFORM DIRECT SUBSTITUTION TO DETERMINE THIS LIMIT.
LET'S GO AHEAD AND DO THAT.
LOOKS LIKE WE'RE GOING TO HAVE 4 + 2,
THAT'S 6 DIVIDED BY 2, WHICH = 3.
THIS LIMIT IS EQUAL TO THREE, WHICH TELLS US THAT
AS WE APPROACH THE POINT (1, 1), THE Z VALUE IS APPROACHING 3.
SO THE HEIGHT OF THIS SURFACE HERE IS APPROACHING 3.
NOW, SINCE THIS SURFACE DOES HAVE DISCONTINUITY,
LET'S TAKE A LOOK AT WHERE THAT OCCURS.
WE'LL TAKE A LOOK AT OUR DENOMINATOR.
IF X AND Y WERE OPPOSITES, THE DENOMINATOR WOULD BE ZERO,
AND WE WOULD HAVE DISCONTINUITY.
SO IF WE CONSIDERED THE PATH WHERE Y = -X--
SO IF WE REPLACE Y WITH -X,
WE'D HAVE A ZERO IN THE DENOMINATOR,
THIS FUNCTION WOULD BE DISCONTINUOUS.
NOW, WHAT I WANT YOU TO NOTICE
IS IF WE ROTATE THIS SURFACE TO LOOK DOWN ON THE XY PLANE,
YOU CAN SEE THE BREAK IN THE SURFACE,
AND THIS BREAK OCCURS ALONG THE LINE Y = -X.
SO WHAT THIS TELLS US IS
IF WE CONSIDERED THE LIMIT AS XY APPROACHED,
LET'S SAY, (1, -1) OR (2, -2),
THE LIMIT WOULD NOT EXIST NOT NECESSARILY
BECAUSE OF THE DISCONTINUITY,
BUT BECAUSE WE CAN SEE GRAPHICALLY
IF WE APPROACH THOSE POINTS ALONG THE LINE Y = -X
FROM DIFFERENT PATHS, THE LIMITS WOULD BE DIFFERENT.
HERE WE HAVE THE LIMIT AS XY APPROACHES (0, 0)
OF THE GIVEN FUNCTION.
I'VE GRAPHED THE SURFACE OVER HERE IN 3D.
LET'S SEE IF WE CAN LOCATE THE POINT (0, 0).
WELL, WHEN X IS ZERO AND Y IS ZERO,
WE'D BE AT THE ORIGIN,
AND THEN FROM THERE WE'D GO UP TO THE SURFACE.
IT'S KIND OF UNCLEAR WHERE THAT POINT WOULD BE.
I DON'T KNOW IF IT'S UP HERE OR DOWN HERE.
LET'S JUST KIND OF PUT A POINT SOMEWHERE HERE IN THE MIDDLE.
WE'RE APPROACHING THAT POINT ON THE SURFACE,
AND JUST FROM THE GRAPH OF THIS SURFACE,
IT IS BEHAVING A LITTLE BIT STRANGE RIGHT AROUND THE ORIGIN.
ONE OF THE THINGS WE NOTICE ON THIS ONE
IS IF WE TRY TO DO DIRECT SUBSTITUTION,
NOTICE WE'RE GOING TO HAVE 0/0,
WHICH IS IN INDETERMINATE FORM.
AND SINCE WE CAN'T PERFORM DIRECT SUBSTITUTION,
WE'RE GOING TO GO AHEAD AND APPROACH THIS
FROM TWO DIFFERENT PATHS.
LET'S SEE WHAT WE DISCOVER.
LET'S CONSIDER THE PATH WHERE Y = X.
WELL, IF Y = X, WE CAN REWRITE THIS LIMIT
AS THE LIMIT AS (X, X) APPROACHES (0, 0).
AGAIN, SINCE Y = X, WE CAN REPLACE Y WITH X.
SO IF WE REPLACE Y WITH X,
THIS WOULD BECOME X x X OR X SQUARED/X SQUARED PLUS--
AND THEN Y SQUARED WOULD BECOME X SQUARED WITH THE SUBSTITUTION.
SO THIS BECOMES THE LIMIT.
THIS BECOMES X SQUARED /2X SQUARED.
WE HAVE A LEADING COEFFICIENT HERE OF ONE,
AND THE REASON I MENTION THAT IS THIS SIMPLIFIES.
AND NOTICE NOW,
THE SIMPLIFIED FUNCTION IS NOT EVEN AFFECTED BY X.
SO THIS LIMIT ALONG THE PATH OF Y = X IS EQUAL TO +1/2.
NOW, LET'S GO AHEAD AND CONSIDER THE PATH Y = -X.
LET'S REWRITE THE LIMIT. WE'LL HAVE THE LIMIT AS X--
NOW, Y = -X, SO WE'LL HAVE (X, -X) APPROACHING (0, 0).
NOW, LET'S PERFORM
THIS SAME SUBSTITUTION INTO THE ORIGINAL FUNCTION.
WE'RE REPLACING Y WITH -X. SO WE'LL HAVE X x -X.
IT'LL BE -X SQUARED.
DENOMINATOR'S GOING TO BE X SQUARED + -X SQUARED.
THAT'S GOING TO BE +X SQUARED. WE'RE GOING TO HAVE +2X SQUARED,
AND THIS SIMPLIFIES JUST LIKE THE LAST ONE.
THE X SQUAREDS SIMPLIFY OUT,
AND NOW THIS LIMIT ALONG THE PATH Y = -X = -1/2.
SINCE THESE TWO LIMITS ARE NOT THE SAME,
WE CAN CONCLUDE THAT THIS LIMIT DOES NOT EXIST.
SO WHAT WE'VE DONE HERE IS APPROACHED (0, 0)
ALONG THIS LINE HERE, Y = X,
AND THEN WE APPROACHED ALONG THE LINE Y = -X,
AND YOU CAN SEE THE YELLOW LIMIT WAS +1/2.
THE GREEN LIMIT WAS -1/2. THEY'RE NOT THE SAME.
THEREFORE, THE LIMIT DOESN'T EXIST.
LET'S TAKE A LOOK AT A FEW MORE.
NOW, ON THIS ONE I JUST WANTED TO SHOW IT'S THE SAME FUNCTION,
BUT NOW, WE'RE APPROACHING THE POINT (4, 2).
SO EVEN THOUGH WE HAD A PROBLEM APPROACHING (0, 0).
WITH THE POINT (4, 2) WE'VE COME OUT FOUR ALONG THE X AXIS
AND THEN OVER TWO ALONG THE Y AXIS.
WE MIGHT BE SOMEWHERE AROUND HERE,
AND NOTICE THE SURFACE
IS NICE AND CONTINUOUS AROUND THIS POINT.
SO WE CAN GO AHEAD AND JUST PERFORM DIRECT SUBSTITUTION
TO DETERMINE THE LIMIT AS XY APPROACHES (4, 2).
THIS WOULD GIVE US 4 x 2 /4 SQUARED + 2 SQUARED.
SO JUST BECAUSE THERE WASN'T A LIMIT APPROACHING (0, 0),
DOESN'T MEAN THERE WON'T BE A LIMIT APPROACHING (4, 2).
SO HERE WE'RE GOING TO HAVE 8 OVER--THIS WOULD BE 16 + 4 = 20.
THIS LIMIT IS GOING TO BE EQUAL TO 2/5.
HERE WE HAVE THE LIMIT AS XY APPROACHES
(0, 0) OF NATURAL LOG X SQUARED + Y SQUARED.
HERE'S THE GRAPH OF THE SURFACE,
AND YOU SEE HERE I'VE ALSO ROTATED IT
SO WE COULD GET A NICE VIEW OF THE Z AXIS.
SO IF I WANTED TO DETERMINE WHERE THE POINT WOULD BE
AS X AND Y APPROACH ZERO,
I WOULD BE APPROACHING THE ORIGIN HERE
FROM ALL DIFFERENT DIRECTIONS.
VERY EASILY SEE THAT AS WE APPROACH THE ORIGIN,
THE SURFACE IS GOING DOWN VERY QUICKLY.
IT'S EVEN MORE OBVIOUS ON THIS GRAPH HERE.
SO IT'S PRETTY EASY TO SEE FROM THIS SURFACE
THAT AS X AND Y APPROACH ZERO,
Z IS APPROACHING NEGATIVE INFINITY,
WHICH MEANS THIS LIMIT DOES NOT EXIST.
ANOTHER WAY TO THINK OF THIS WOULD BE TO CONSIDER
THE GRAPH OF A FUNCTION IN TWO VARIABLES,
WHICH WE SHOULD BE FAMILIAR WITH.
THIS GRAPH LOOKS SOMETHING LIKE THIS,
AND WE KNOW THAT AS X APPROACHES ZERO FROM THE RIGHT,
THIS FUNCTION VALUE APPROACHES NEGATIVE INFINITY.
SO FOR THE SAME REASON,
X SQUARED + Y SQUARED IS GOING TO APPROACH ZERO
WHEN X AND Y APPROACH ZERO.
THEREFORE, Z IS GOING TO APPROACH NEGATIVE INFINITY.
LET'S TAKE A LOOK AT THIS LAST ONE.
HERE WE HAVE THE LIMIT AS XY APPROACHES (2, 4).
IF WE TAKE A LOOK AT THE GRAPH OF THIS SURFACE,
IT MIGHT BE KIND OF SURPRISING.
IT LOOKS LIKE THIS IS JUST A PLANE,
AND THIS DEFINITELY
DOES NOT LOOK LIKE AN EQUATION FOR A PLANE.
AND IF WE TRY TO PERFORM DIRECT SUBSTITUTION,
OUR DENOMINATOR IS GOING TO EQUAL ZERO.
SO IT'S UNDEFINED AT THE POINT (2, 4),
BUT THIS RATIONAL FUNCTION CAN ACTUALLY BE SIMPLIFIED.
THE NUMERATOR IS FACTORABLE
BECAUSE THERE'S A COMMON FACTOR OF Y.
SO LET'S GO AHEAD AND FACTOR THE NUMERATOR.
SO WE'LL HAVE Y x Y - 2X/Y - 2X,
AND NOTICE THAT THIS RATIONAL FUNCTION SIMPLIFIES NICELY,
AND WHEN IT SIMPLIFIES, IT ELIMINATES THAT DISCONTINUITY.
SO NOW, WE JUST HAVE THE LIMIT AS XY APPROACHES (2, 4) OF Y,
AND Y IS APPROACHING 4. SO THIS LIMIT IS EQUAL TO 4.
SO JUST LIKE WE SAW WITH FUNCTIONS
IN TERMS OF ONE VARIABLE,
IF THE RATIONAL FUNCTION COULD BE SIMPLIFIED,
IT CAN OFTEN REMOVE THE DISCONTINUITY,
MAKING THE LIMIT MUCH EASIER TO DETERMINE.
OKAY. I HOPE YOU FOUND THESE EXAMPLES HELPFUL.
THANK YOU FOR WATCHING.