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NCERT Class XI Mathematics - Limits and Derivatives - Solutions

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Question : 18
Total: 72
lim
x→0
ax+xcosx
bsinx
Solution:  
We have,
lim
x→0
ax+xcosx
bsinx
=
lim
x→0
x(a+cosx)
bsinx
 =
lim
x→0
x
sinx
(a+cosx)
b
= (
lim
x→0
x
sinx
).(
lim
x→0
a+cosx
b
) = (1) .
a+cos0
b
 =
a+1
b
.
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