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- NOW WE'LL LOOK AT TWO EXAMPLES,
DETERMINING LIMITS AT INFINITY WHERE THE FUNCTIONS
INVOLVE SQUARE ROOTS.
WE KNOW FROM THE PREVIOUS EXAMPLES
WHEN WE HAD A RATIONAL FUNCTION, WE ALWAYS DIVIDED EVERY TERM
BY THE HIGHEST POWER OF X FROM THE DENOMINATOR.
BUT IN THIS CASE, SINCE OUR FUNCTION
INVOLVES A SQUARE ROOT, WE HAVE TO BE EXTREMELY CAREFUL
ABOUT THE SIGNS.
SO INSTEAD OF DIVIDING BY THE HIGHEST POWER OF X
FROM THE DENOMINATOR, WE'RE GOING TO FACTOR INSTEAD.
BUT BEFORE WE DO THIS, THE REASON WE HAVE TO BE
SO CAREFUL ABOUT DETERMINING THESE LIMITS
IS BECAUSE IF WE HAVE THE SQUARE ROOT OF X SQUARED,
THIS ISN'T EQUAL TO JUST X.
THIS HAS TO BE A POSITIVE VALUE.
THEREFORE IT'S EQUAL TO THE ABSOLUTE VALUE OF X,
WHICH MEANS THE ABSOLUTE VALUE OF X,
IF WE WANT TO WRITE THIS IN TERMS OF X, IS EQUAL TO X
IF X IS GREATER THAN OR EQUAL TO 0,
BUT IT WOULD BE -X IF X IS LESS THAN 0,
MEANING IF X IS NEGATIVE, WE CAN WRITE THE ABSOLUTE VALUE
OF X AS THE OPPOSITE OF X.
AND WE'LL HAVE TO DO THIS WHEN WE FACTOR
TO DETERMINE THESE LIMITS.
SO SINCE OUR DENOMINATOR IS 4X + 3,
WE'RE GOING TO FACTOR OUT X FROM BOTH THE NUMERATOR
AND DENOMINATOR.
SO LET'S START WITH THE DENOMINATOR.
WE'LL HAVE THE LIMIT AS X APPROACHES INFINITY.
WE FACTOR OUT X FROM THE DENOMINATOR,
WE WOULD HAVE X TIMES THE QUANTITY 4 + 3 DIVIDED BY X.
NOTICE HOW X TIMES 3 DIVIDED BY X IS STILL 3.
NOW WE WANT TO FACTOR OUT X FROM THE SQUARE ROOT,
BUT WE KNOW IF WE FACTOR OUT AN X FROM A SQUARE ROOT
IT WOULD ACTUALLY BE THE ABSOLUTE VALUE OF X,
WHICH IN THIS CASE, SINCE X IS APPROACHING POSITIVE INFINITY,
OR X IS GREATER THAN OR EQUAL TO 0,
WE CAN JUST FACTOR OUT X.
BUT IF WE FACTOR OUT X, REMEMBER,
UNDERNEATH THE SQUARE ROOT, IT WOULD ACTUALLY BE X SQUARED.
SO IF WE FACTOR OUT X SQUARED UNDERNEATH THE SQUARE ROOT
FROM 5, WE WOULD HAVE 5 DIVIDED BY X SQUARED.
IF WE FACTOR OUT X SQUARED FROM 4X SQUARED
UNDERNEATH THE SQUARE ROOT, WE WOULD JUST HAVE PLUS 4.
NOW THAT WE HAVE THIS IN FACTOR FORM,
X/X SIMPLIFIES TO 1.
AND NOW WE'LL SEE WHAT HAPPENS TO EACH TERM
AS X APPROACHES INFINITY.
WELL, 5 DIVIDED BY X SQUARED WOULD APPROACH 0
AS X APPROACHES INFINITY.
NOTICE HOW THIS 4 AND THIS 4 ARE NOT AFFECTED BY X,
AND 3 DIVIDED BY X WOULD APPROACH 0.
SO WE'RE LEFT WITH A NUMERATOR OF THE SQUARE ROOT 4
AND A DENOMINATOR OF 4.
WELL, THE SQUARE ROOT OF 4 = 2, 2/4 = 1/2.
SO THIS FIRST LIMIT IS EQUAL TO 1/2.
NOW, LET'S CONSIDER THE SECOND LIMIT,
AND NOTICE THE ONLY DIFFERENCE HERE
IS THAT X IS APPROACHING NEGATIVE INFINITY.
SO WE'LL START AGAIN BY FACTORING OUT X
FROM THE DENOMINATOR.
SO WE'D HAVE X TIMES THE QUANTITY 4 + 3 DIVIDED BY X.
NOW WE WANT TO FACTOR OUT X FROM THE SQUARE ROOT,
BUT, AGAIN, WE KNOW THAT VALUE HAS TO BE POSITIVE.
AND NOW X IS APPROACHING NEGATIVE INFINITY,
SO THE ABSOLUTE VALUE OF X WOULD BE EQUAL TO -X
THIS TIME BECAUSE X IS LESS THAN 0.
SO IF WE WANT TO FACTOR OUT X, WE'D HAVE TO FACTOR OUT -X.
AGAIN, BECAUSE X IS APPROACHING NEGATIVE INFINITY,
THEREFORE THIS IS NOW A POSITIVE VALUE.
AND THEN WE'RE STILL LEFT WITH, UNDERNEATH THE SQUARE ROOT,
THIS WOULD STILL BE X SQUARED.
SO WE'D HAVE 5 DIVIDED BY X SQUARED + 4.
NOW THAT IT'S IN FACTORED FORM, NOTICE HOW -X DIVIDED BY X
SIMPLIFIES TO -1.
5 DIVIDED BY X SQUARED STILL APPROACHES 0,
SO DOES 3 DIVIDED BY X.
AND THIS 4 AND THIS 4 ARE NOT AFFECTED BY X,
SO THE ONLY DIFFERENCE HERE IS THAT WE WOULD HAVE
NEGATIVE SQUARE ROOT 4 DIVIDED BY 4,
WHICH WOULD SIMPLIFY TO -1/2.
SO AS YOU CAN SEE, WE HAVE TO BE EXTREMELY CAREFUL
WHEN DETERMINING THE SIGN OF OUR LIMIT IF THE FUNCTION
INVOLVES A SQUARE ROOT.
NOW, LET'S GO AHEAD AND GRAPH OUR FUNCTION
TO VERIFY THESE TWO LIMITS.
NOTICE AS X APPROACHES POSITIVE INFINITY IT DOES APPEAR
THE FUNCTION VALUE IS APPROACHING 1/2.
AND AS X APPROACHES NEGATIVE INFINITY IT DOES APPEAR
THE FUNCTION VALUE IS APPROACHING -1/2.
THIS ALSO VERIFIES THAT OUR FUNCTION
HAS TWO HORIZONTAL ASYMPTOTES.
ONE IS Y = 1/2 AND ANOTHER THAT'S Y = -1/2.
SO WE HAVE A HORIZONTAL ASYMPTOTE
OF Y = 1/2 TO THE RIGHT AND Y = -1/2 TO THE LEFT.
OKAY, I HOPE THIS EXPLANATION WAS HELPFUL.