Page 71 - Calculus Workbook For Dummies

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Chapter 4: Nitty-Gritty Limit Problems

s lim x x = 3
x " 3 ! x
According to the “big” over “small” tip, this answer must be infinity. Or you can get this result with
your calculator. If you set the table (don’t forget: fork on the left, spoon on the right) the same as
in problem 19, you’ll see “undef” for all y values. You’ve got to be careful when trying to interpret
what “undef” (for “undefined”) means on your calculator, but in this case — because a 0 denomina-
tor is impossible and thus can’t be the reason the fraction is undefined — it looks like “undef”
means infinity. To confirm this, make tblStart and ∆tbl smaller, say, 10. Sure enough, the y values
grow huge very fast, and you can safely conclude that the limit is infinity.
t lim 5 x + 2 = 5
2
x " 3 4 x - 1 2
1. Divide numerator and denominator by x.
5 x + 2
x
= lim
2
x " 3 4 x - 1
x
2
2. Put the x into the square root (it becomes x ).
5 x + 2
x
= lim
2
x " 3 4 x - 1
x 2
3. Distribute the division.
2
5 + x
= lim
x " 3 1
4 - 2
x
4. Plug in and simplify.
2
5 + 3
=
1
4 -
3 2
5 + 0
=
4 - 0
5
=
2
*u lim 4 + 16 x - 3 j 3
2
x =
x
`
x " - 3 8
1. Put the entire expression over 1 so you can use the conjugate trick.
2
2
4 ` x + 16 x - 3 xj 4 ` x - 16 x - 3 xj
= lim $
2
x " - 3 1 4 ` x - 16 x - 3 xj
2. FOIL the numerator.
16 x - _ 16 x - 3 xi
2
2
= lim
2
x " - 3 4 x - 16 x - 3 x
2
3. Simplify the numerator and factor 16x inside the radicand.
3 x
= lim
x " - 3 2 3
4 x - 16 c m
x 1 -
16 x