Solve it - limx - 0 cos ax - cos bx-cos cx - cos dx

lim_x→ 0cos ax cos bxcos cx cos dx Apply L Hospital Rule two times. = lim_x→ 0 asin ax+bsin bx csin cx+dsin dx = lim_x→ 0 a^2cos ax+b^2cos bx c^2c ...

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Standard XII

Mathematics

Theorems for Differentiability

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Question

Solve it :-
$lim x \to 0 \frac{cos a x - cos b x}{cos c x - cos d x}$

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Solution

$lim x \to 0 \frac{cos a x - cos b x}{cos c x - cos d x}$
Apply L-Hospital Rule two times.
$= lim x \to 0 \frac{- a sin a x + b sin b x}{- c sin c x + d sin d x}$
$= lim x \to 0 \frac{- a^{2} cos a x + b^{2} cos b x}{- c^{2} cos c x + d^{2} cos d x}$ .
$= \frac{b^{2} - a^{2}}{d^{2} - c^{2}}$

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Solve it :-limx→0cosax−cosbxcoscx−cosdx

limx→0cosax−cosbxcoscx−cosdxApply L-Hospital Rule two times.=limx→0−asinax+bsinbx−csincx+dsindx=limx→0−a2cosax+b2cosbx−c2coscx+d2cosdx.=b2−a2d2−c2

Solve it :-
$lim x \to 0 \frac{cos a x - cos b x}{cos c x - cos d x}$

$lim x \to 0 \frac{cos a x - cos b x}{cos c x - cos d x}$
Apply L-Hospital Rule two times.
$= lim x \to 0 \frac{- a sin a x + b sin b x}{- c sin c x + d sin d x}$
$= lim x \to 0 \frac{- a^{2} cos a x + b^{2} cos b x}{- c^{2} cos c x + d^{2} cos d x}$ .
$= \frac{b^{2} - a^{2}}{d^{2} - c^{2}}$