_ x 0 (15^x-3^x-5^x+1 ^2 x )= - Vidyadip

MCQ

$\lim _{x \rightarrow 0}\left(\frac{15^x-3^x-5^x+1}{\sin ^2 x}\right)=$

A
$\log 15$
B
$\log 3+\log 5$
✓
$\log 3 . \log 5$
D
$3 \log 5$

✓

Answer

Correct option: C.

$\log 3 . \log 5$

(C) $\log 3 . \log 5$
Hint:
$\lim _{x \rightarrow 0}\left(\frac{15^x-3^x-5^x+1}{\sin ^2 x}\right)$
$=\lim _{x \rightarrow 0}\left[\frac{\left(5^x-1\right)\left(3^x-1\right)}{x^2} \times \frac{x^2}{\sin ^2 x}\right] $
$\because x \rightarrow 0, x \neq 0$
$\therefore x^2 \neq 0$
$=\log 3 \cdot \log 5$

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