Question

Evaluate:
$\rm \displaystyle \lim_{x \to 0}\ \frac{a^{x}+b^{x}+c^{x}-3}{\tan x}$

Answer 1

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Answer 2

Answer:

$\rm \displaystyle \lim_{x \to 0}\ \frac{a^{x}+b^{x}+c^{x}-3}{\tan x} = log(abc)$

Step-by-step explanation:

Since we need to evaluate the $\rm \displaystyle \lim_{x \to 0}\ \frac{a^{x}+b^{x}+c^{x}-3}{\tan x}$

Apply $\lim_{x\to0}$ in given question we get

$\frac{a^{0}+b^{0}+c^{0} -3 }{tan0}$ = $\frac{1+1+1-3}{0}$ = $\frac{0}{0}$

So , apply L'Hospital rule (Differentiate each term with respect to the variable) we get ,

$\rm \displaystyle \lim_{x \to 0}\ \frac{a^{x}loga+b^{x}logb+c^{x}logc}{(secx)^{2} }$

Apply limit x tends to 0 we get,

$\frac{loga+logb+logc}{1} =log(abc)$

So,

$\rm \displaystyle \lim_{x \to 0}\ \frac{a^{x}+b^{x}+c^{x}-3}{\tan x} = log(abc)$