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NCERT Class XI Mathematics - Limits and Derivatives - Solutions

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Question : 20
Total: 72
lim
x→0
sinax+bx
ax+sinbx
 , a , b , a + b ≠ 0
Solution:  
We have,
lim
x→0
sinax+bx
ax+sinbx
Dividing the numerator and denominator by x, we get
lim
x→0
sinax
x
+b
a+
sinbx
x
 =
lim
x→0
sinax
ax
.a+b
a+
sinax
bx
.b
 =
a[
lim
x→0
sinax
ax
]+b
a+[
lim
x→0
sinbx
bx
].b
=
a(1)+b
a+b.(1)
 =
a+b
a+b
 = 1.
Since
lim
θ→0
sinθ
θ
 = 1.
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