as x⟶0. To solve this, we will use the rationalization method. Multiplying both numerator and denominator by the conjugate of the numerator, we get: ‌
lim
x⟶0
‌
√a+x−√a
x√a(a+x)
*‌
√a+x+√a
√a+x+√a
‌
lim
x⟶0
‌
(√a+x)2−(√a)2
x√a(a+x)(√a+x+√a)
‌
lim
x⟶0
‌
a+x−a
x√a(a+x)(√a+x+√a)
‌
lim
x⟶0
‌
x
x√a(a+x)(√a+x+√a)
‌
lim
x⟶0
‌
1
√a(a+x)(√a+x+√a)
Now, we can directly substitute x=0 into the expression, to get: ‌‌