The value of limx- 01-cosx22x4 is

Lim_x→ 01 cosx^22x^4=lim_x→ 02sin^2x^2/22x^4 =lim_x→ 0 ( sinx^2/22x^2/2 )^2=14 Alternative Solution: lim_x→ 01 cosx^22x^4 ( 00 ) Apply L ' Hospital Rule, ...

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Standard XII

Mathematics

Right Hand Limit

The value of ...

Question

The value of ${lim}_{x \to 0} \frac{1 - c o s (x^{2})}{2 x^{4}}$ is

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\frac{1}{2}

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\frac{1}{4}

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Solution

The correct option is C $\frac{1}{4}$
${lim}_{x \to 0} \frac{1 - c o s x^{2}}{2 x^{4}} = {lim}_{x \to 0} \frac{2 s i n^{2} x^{2} / 2}{2 x^{4}}$

$= {lim}_{x \to 0} {(\frac{s i n x^{2} / 2}{2 x^{2} / 2})}^{2} = \frac{1}{4}$

Alternative Solution:

${lim}_{x \to 0} \frac{1 - c o s x^{2}}{2 x^{4}} (\frac{0}{0})$

Apply L ' Hospital Rule,

$= {lim}_{x \to 0} \frac{s i n x^{2} .2 x}{8 x^{3}}$

$= {lim}_{x \to 0} \frac{s i n x^{2}}{4 x^{2}} = \frac{1}{4}$

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The value of limx→01−cos(x2)2x4 is

The correct option is C 14limx→01−cosx22x4=limx→02sin2x2/22x4 =limx→0(sinx2/22x2/2)2=14 Alternative Solution: limx→01−cosx22x4(00) Apply L ' Hospital Rule, =limx→0sinx2.2x8x3 =limx→0sinx24x2=14

The value of ${lim}_{x \to 0} \frac{1 - c o s (x^{2})}{2 x^{4}}$ is