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Select the correct answer from the given alternatives.

\(\lim\limits_{x\longrightarrow0}[\frac{15^x-3^x-5^x+1}{sin^2x}]=\)

lim [15x-3x-5x+1/sin2x](x ∈ 0) =

(A) log 15 

(B) log 3 + log 5

(C) log 3 . log 5 

(D) 3 log 5

1 Answer

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Best answer

Correct option is: (C) log 3. log 5

\(\lim\limits_{x\longrightarrow0}[\frac{15^x-3^x-5^x+1}{sin^2x}]\) 

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