http://www.ck12.org Chapter 14. Concepts of Calculus

Solution:

xlim→ 3 (x(√−x^3 −)(x√+ 33 ))·

(√x+√ 3 )

(√x+√ 3 )=xlim→ 3 (x−^3 )(x+^3 )

(√x+√ 3 )

(x− 3 )

=xlim→ 3 (x+ 3 )

(√

x+

√

3

)

= 6 · 2 √ 3

= 12

√

3

Example B
Evaluate the following limit: limx→ 25 √x−x^25 − 5.

Solution:

xlim→ 25 √x−x−^255 =xlim→ (^25) ((√x−x−^255 ))·(
√x+ 5 )
(√x+ 5 )
=xlim→ 25 (x−^25 )(
√x+ 5 )
(x− 25 )
=xlim→ 25 (√x+ 5 )

√

25 + 5

= 10

Example C

Evaluate the following limit: limx→ 7

√x+ 2 − 3
x− 7.
Solution:

xlim→ 7

√x+ 2 − 3

x− 7 =limx→ 7

(√x+ 2 − 3 )

(x− 7 ) ·

(√x+ 2 + 3 )

(√x+ 2 + 3 )

=xlim→ (^7) (x− 7 ()x·+(√^2 x−+^9 ) 2 + 3 )
=xlim→ (^7) (x− 7 )·((x√−x^7 )+ 2 + 3 )
=limx→ 7 (√x+^12 + 3 )
= √ 7 +^12 + 3
=^16
Concept Problem Revisited
In order to evaluate the limit of the following rational expression, you need to multiply by a clever form of 1 so that
when you substitute there is no longer a zero factor in the denominator.