Fx- cos x x 1 2 sin x x2 2x tan x x 1 then lim x- 0 fx -

The correct option is A 0 f(x)= cos x amp; x amp; 1 2sin x amp; x^ 2 amp; 2x tan x amp; x amp; 1 R _ 3 → R _ 3 R ...

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Determinant

fx= cos x ...

Question

$f (x) =$ $∣ ∣ ∣ ∣ \begin{matrix} cos x & x & 1 2 sin x & x^{2} & 2 x tan x & x & 1 \end{matrix} ∣ ∣ ∣ ∣$ then $lim x \to 0 f (x) =$

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Solution

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Similar questions

f (x) = ∣ ∣ ∣ ∣ \begin{matrix} cos x & x & 1 2 sin x & x^{2} & 2 x tan x & x & 1 \end{matrix} ∣ ∣ ∣ ∣

lim x \to 0 \frac{f (x)}{x}

f (x) = ∣ ∣ ∣ ∣ \begin{matrix} cos x & x & 1 2 sin x & x^{2} & 2 x tan x & x & 1 \end{matrix} ∣ ∣ ∣ ∣

{lim}_{x \to 0} \frac{f^{'} (x)}{x} =

f (x) = ∣ ∣ ∣ ∣ \begin{matrix} c o s x & x & 1 2 s i n x & x^{2} & 2 x t a n x & x & 1 \end{matrix} ∣ ∣ ∣ ∣

l i m x ⟶ 0 \frac{f^{^{'}} (x)}{x} =

f (x) = ∣ ∣ ∣ ∣ \begin{matrix} cos x & x & 1 2 sin x & x^{2} & 2 x tan x & x & 1 \end{matrix} ∣ ∣ ∣ ∣

lim x \to 0 \frac{f^{'} (x)}{x}

f (0) = 1

f^{'} (0) = 9

{lim}_{x \to 0} \frac{f (x) - f (\frac{x}{2}) + f (\frac{x}{2^{2}}) - f (\frac{x}{2^{3}})}{f (x) - 1}

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