http://www.ck12.org Chapter 6. Transcendental Functions
limx→ 0√ 2 +x−√ 2
x =limx→ 0[d
dx(√ 2 +x−√ 2 )dxd(x)]
=
[ 1 /( 2 √ 2 +x)
1]
x= 0
= 2 √^12=√ 2
4.
Example 2:
Find limx→ 01 −x 2 cos 2+ 2 xx
Solution:
We can see that the limit is 0/0 whenx=0 is substituted.
Using L’Hospital’s rule,
xlim→ 01 x−^2 +cos 2 2 xx=[2 sin 2x
2 x+ 2]
x= 0
= 0 / 2
= 0.Example 3:
Use L’Hospital’s rule to evaluate limx→ 3 xx^2 −− 39.
Solution:
limx→ 3 =x(^2) − 9
x− 3 =limx→ 3
2 x
1 =^6.
Example 4:
Evaluate limx→ 0 sin 3xx.
Solution:
limx→ 0 sin 3xx=limx→ 0 3 cos 3 1 x= 3.
Example 5:
Evaluate limx→π/ 25 −cos5 sinxx.
Solution:
We can use L’Hospital’s rule since the limit produces the 0/0 oncex=π/2 is substituted. Hence
x→limπ/ 25 −cos5 sinx x=x→limπ/ 20 −−5 cossinxx=−^01 =^0.