Mathematics Assignment: L'Hospital Rule with Solution

Mathematics Assignment
Student Name:
Instructor Name:
Course Number:
12th December 2019
Q1a)

lim
x→ 0+¿ ( x2 lnx ) ¿
¿
We know that a . b= a
1
b
where b≠ 0
( x2 lnx )= lnx
1
x2
lim
x→ 0+¿ ( x2 lnx )= lim
x→ 0+ ¿
(lnx
1
x2 )¿
¿¿
¿
lim
x→ 0+¿
( lnx
1
x2 ) ¿
¿
By applying the L’Hospital rule we get that
lim
x→ 0+¿
( lnx
1
x2 )= lim
x→ 0+ ¿
( 1
x
−2
x3 ) ¿
¿¿
¿
lim
x→ 0+¿
( 1
x
−2
x3 )= lim
x→ 0+ ¿( −x2
2 ) ¿
¿ ¿
¿
Substituting x=0 in (−x2
2 ) we shall have
( −02
2 ) =0
lim
x→ 0+¿ ( x2 lnx ) =0 ¿
¿
b)
lim
x→−∞
( x2 + √ x2+ x )
We shall multiply ( x2 + √ x2 + x ) by its conjugate i.e. x2− √ x2+ x
( x2+ √ x2 + x )( x2− √x2 + x )
( x2− √ x2 + x ) =¿ ¿

lim
x→−∞
( x2 + √ x2+ x ) = lim
x →−∞ ( −x
( x2− √ x2 + x ) )
lim
x→−∞ ( −x
( x2− √ x2 + x ) ) =− lim
x →−∞ ( x
( x2− √ x2 + x ) )
We shall divide x
( x2− √ x2 + x ) by the highest power of the denominator to get
1
( 1+ √ 1+ 1
x )
− lim
x→−∞ ( x
( x2− √ x2 + x ) ) =− lim
x →−∞
( 1
( 1+ √ 1+ 1
x ) )
− lim
x →−∞
(1)
lim
x →−∞ ( 1+ √ 1+ 1
x ) = −1
1+ √ 1 =−1
2
lim
x→−∞
( x2 + √ x2+ x )=−1
2
c)
lim
x→ B
¿ ¿
= lim
x→ B ( ( x−B)
( x−B ) ( x+ 1) ) =lim
x → B ( 1
( x+ 1) )
Substituting x=B in ( 1
(x +1) ) gives
( 1
( B+1) )
lim
x→ B
¿ ¿
d)
lim
x→ ∞ ( 1
1.2 .3 + 1
2.3 .4 + 1
3.4 .5 + ...+ 1
x ( x+ 1 ) ( x+ 2) )

¿ lim
x→ ∞ ( 1
6 + 1
24 + 1
60 + ...+ 1
x ( x+ 1 ) ( x+2) )
=lim
x→ ∞ ( 9
40 + ...+ 1
x ( x+ 1 ) (x+ 2) )
Substituting x=∞ in 1
x ( x +1 ) ( x+ 2) gives
1
∞ ( ∞+1 )( ∞+ 2)= 1
∞ =0
lim
x→ ∞ ( 9
40 + ...+ 1
x ( x+ 1 ) (x+ 2) ) = 9
40 + ...+0
As we continue with the series the terms decrease towards zero since the
denominator of each term increases. This in turn results into the sum of such terms
having no effect on our answer. Thus
lim
x→ ∞ ( 9
40 + ...+ 1
x ( x+ 1 ) (x+ 2) )= 9
40 + ...+0= 9
40 + 0+0= 9
40
lim
x→ ∞ ( 1
1.2 .3 + 1
2.3 .4 + 1
3.4 .5 + ...+ 1
x ( x+ 1 ) (x+ 2) ) = 9
40
Q2) a
f (x)=
{ 2 x2−x −6
x−2 if x <2
6 if x=2
7 x
2 if x >2
i) f (x) is defined at x=2 since f ( 2 ) =6
ii)f ( x )= 2 x2 −x−6
x−2 = ( 2 x +3 )( x−2)
( x−2)
f ( x )=2 x +3
lim
x →2
( 2 x +3 )=2 ( 2 ) +3=7