Limx- 0xcos x-2sin xx2-tan x-

lim _ x→ 0 xcos #160; x+2sin #160; x x^ 2 +tan #160; x . #160; (L Hopital rule) cos x xsin x+2cos x2x+^2x 1 0+20+1=3 ...

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Standard XII

Mathematics

Derivative of Standard Functions

limx→ 0xcos x...

Question

$lim x \to 0 \frac{x cos x + 2 sin x}{x^{2} + tan x}$ .

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Solution

${lim}_{x \to 0} \frac{x cos x + 2 sin x}{x^{2} + tan x}$ . (L-Hopital rule)
$\frac{cos x - x sin x + 2 cos x}{2 x + {sec}^{2} x}$
$\frac{1 - 0 + 2}{0 + 1} = 3$

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\lim_{x \to 0} \frac{x \cos x + 2 \sin x}{x^{2} + \tan x}

lim x \to 0 \frac{x cos x - log (1 + x)}{x^{2}}

equals

You are given
$cos x = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \frac{x^{6}}{6!} . . . . . .$ ;
$sin x = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} . . . . . .$ ;
$tan x = x + \frac{x^{3}}{3} + \frac{2 x^{5}}{15} . . . . . .$
Then the value of
$lim x \to 0 \frac{x cos x + sin x}{x^{2} + tan x}$ is

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limx→0xcosx+2sinxx2+tanx.

limx→0xcosx+2sinxx2+tanx. (L-Hopital rule)cosx−xsinx+2cosx2x+sec2x1−0+20+1=3

$lim x \to 0 \frac{x cos x + 2 sin x}{x^{2} + tan x}$ .

${lim}_{x \to 0} \frac{x cos x + 2 sin x}{x^{2} + tan x}$ . (L-Hopital rule)
$\frac{cos x - x sin x + 2 cos x}{2 x + {sec}^{2} x}$
$\frac{1 - 0 + 2}{0 + 1} = 3$