Tip:
Highlight text to annotate it
X
- NOW WE'LL LOOK AT TWO EXAMPLES
TO SHOW HOW FACTORING CAN BE HELPFUL
WHEN DETERMINING LIMITS.
NOTICE IN BOTH CASES WE HAVE THE LIMITS OF RATIONAL FUNCTIONS.
FIRST WE HAVE THE LIMIT AS X APPROACHES 2
(X SQUARED - X) - 2 DIVIDED BY (X SQUARED + X) - 6.
BEFORE WE FACTOR THIS THOUGH,
LET'S LOOK AT THE GRAPH OF THIS FUNCTION.
NOTICE HOW THE RATIONAL FUNCTION HAS A HOLE AT X = 2
OR IT HAS POINT DISCONTINUITY
AND IT HAS INFINITE DISCONTINUITY
OR A VERTICAL ASYMPTOTE AT X = -3.
SO BECAUSE THIS FUNCTION IS UNDEFINED AT X = 2,
WE COULD NOT PERFORM DIRECT SUBSTITUTION
TO DETERMINE THIS LIMIT.
WE'D HAVE A 0 IN THE DENOMINATOR.
SO LET'S GO AHEAD AND FACTOR THE NUMERATOR AND DENOMINATOR
TO BETTER UNDERSTAND THE GRAPH OF THIS FUNCTION.
WE'D HAVE THE LIMIT AS X APPROACHES 2
AND THEN THE NUMERATOR
IS GOING TO FACTOR INTO 2 BINOMIAL FACTORS
AND SO WILL THE DENOMINATOR.
SO TO FACTOR (X SQUARED - X) - 2 WE'LL HAVE X AND X.
THE FACTORS OF -2 THAT ADD TO -1 WOULD BE -2 AND +1
AND THEN FOR OUR DENOMINATOR WE'D HAVE X AND X
AND THE FACTORS OF -6 THAT ADD TO +1 WOULD BE -2 AND +3.
SO NOTICE HOW WE HAVE THIS COMMON FACTOR OF X - 2
AND THE 0'S OF THE COMMON FACTORS
PRODUCE HOLES IN THE GRAPH AS WE SEE HERE.
REMEMBER HAVING A HOLE HERE AT X = 2,
DOES NOT MEAN A LIMIT DOES NOT EXIST.
IN FACT, GRAPHICALLY WE CAN SEE AS WE APPROACH +2
FROM THE RIGHT SIDE OR POSITIVE SIDE,
AND AS WE APPROACH +2 FROM THE LEFT SIDE OR NEGATIVE SIDE,
WE ARE APPROACHING THE SAME FUNCTION VALUE
AND THEREFORE THIS LIMIT DOES EXIST.
BUT BECAUSE WE HAVE THE FUNCTION IN FACTORED FORM NOW,
WE CAN ACTUALLY SIMPLIFY OUT THE COMMON FACTOR HERE
WHICH WOULD GIVE US THE SAME GRAPH
WITHOUT THE HOLE AND THEREFORE IN THE SIMPLIFIED FORM
WE CAN PERFORM DIRECT SUBSTITUTION
TO DETERMINE THIS LIMIT.
THIS LIMIT WOULD BE EQUAL TO THE LIMIT AS X APPROACHES 2
OF THE SIMPLIFIED FUNCTION OF (X + 1) DIVIDED BY (X + 3).
SO IF WE PERFORM DIRECT SUBSTITUTION NOW,
WE WOULD HAVE JUST 2 + 1 DIVIDED BY 2 + 3 OR 3/5.
OUR LIMIT IS EQUAL TO 3/5 MEANING AS WE APPROACH X = 2
FROM THE LEFT AND THE RIGHT,
WE ARE APPROACHING THE FUNCTION VALUE OF 3/5.
REMEMBER THE FUNCTION VALUES ARE THE Y VALUES.
NOW LOOKING AT OUR SECOND EXAMPLE,
LET'S GO AHEAD AND JUMP TO THE GRAPH.
NOTICE HOW THE FUNCTION IS UNDEFINED AT X = -2,
BUT ONCE AGAIN WE'D HAVE DIVISION BY 0.
BUT LOOKING AT THE GRAPH, NOTICE HERE WHERE X = -2,
WE CAN GRAPHICALLY SEE THE LIMIT DOES EXIST AS WE APPROACH -2
FROM THE POSITIVE SIDE OR RIGHT SIDE,
AND AS WE APPROACH -2 FROM THE NEGATIVE SIDE OR LEFT SIDE,
WE CAN SEE FROM THE GRAPH
WE ARE APPROACHING THE FUNCTION VALUE OF 12.
TO FIND THIS LIMIT ALGEBRAICALLY,
ONCE AGAIN WE'LL FACTOR THE NUMERATOR AND DENOMINATOR.
SO WE HAVE THE LIMIT AS X APPROACHES -2.
NOW THE DENOMINATOR DOES NOT FACTOR.
WE HAVE X + 2 BUT THE NUMERATOR IS A SUM OF CUBES,
WHICH WE CAN FACTOR USING THIS FORMULA HERE.
SO IN OUR CASE "A" WOULD BE X AND B WOULD BE 2,
SINCE 2 CUBED IS 8.
SO WE WOULD HAVE (X + 2),
AGAIN FOR OUR PURPOSES "A" = X AND B = 2.
SO FOR "A" SQUARED WE'D HAVE X SQUARED AND THEN -A x B,
THAT WOULD BE -2 x X OR -2X
AND THEN +B SQUARED THAT WOULD BE +2 SQUARED, OR +4.
NOTICE IN FACTORED FORM WE DO HAVE A COMMON FACTOR OF X + 2.
NOTICE THE 0 OF THESE 2 FACTORS IS X = -2
WHICH IS WHERE THE HOLE OCCURS.
SO NOW WE CAN SIMPLIFY OUT THIS COMMON FACTOR
AND USE THE SIMPLIFIED FUNCTION
TO DETERMINE THIS LIMIT ALGEBRAICALLY.
THIS LIMIT IS EQUAL TO THE LIMIT
AS X APPROACHES -2 OF X SQUARED - 2X + 4,
PERFORMING DIRECT SUBSTITUTION,
WE WOULD HAVE -2 SQUARED - 2 x (-2 + 4)
WHICH WOULD JUST BE 4 + 4 + 4 WHICH IS EQUAL TO 12.
SO WHEN DETERMINING LIMITS OF RATIONAL FUNCTIONS,
IT'S OFTEN HELPFUL TO FACTOR AND SIMPLIFY WHICH THEN OFTEN
ALLOWS DIRECT SUBSTITUTION TO DETERMINE THE LIMIT.
I HOPE YOU FOUND THIS HELPFUL.