Explanation
1. Limit Setup:
Humein niche di gayi limit solve karni hai:
L=x→0limxln(1+ax)−ln(1−bx)
2. Properties ka upyog:
Hum jaante hain ki logarithm ki ek standard limit hoti hai:
Ab hum apne function ko do hisson mein todte hain:
L=x→0lim[xln(1+ax)−xln(1−bx)]
3. Adjusting terms:
Pehle term ko a se multiply aur divide karenge, aur dusre term ko −b se:
L=x→0lim[a⋅axln(1+ax)−(−b)⋅−bxln(1−bx)]
4. Limit apply karne par:
Kyunki axln(1+ax)→1 aur −bxln(1−bx)→1 jab x→0:
Alternate Method (L'Hopital's Rule):
Chunki x=0 rakhne par ye 00 form ban raha hai, hum numerator aur denominator ko differentiate kar sakte hain:
x→0limdxd[x]dxd[ln(1+ax)−ln(1−bx)]
Ab x=0 rakhne par:
Final Answer:
Function ko x=0 par continuous hone ke liye, f(0) ki value honi chahiye:
f(0)=a+b
