Let - fx- begin cases lendcases -If f x is continuous at x -0- then k equalsA- 1B- 4-6C- t-12D- 8

The correct option is C 112k=lim_x → 0e^ x^22 cos xxln(1+x)sin x(e^x 1) k=lim_x → 0e^ x^22 cos xx^4 =lim_x → 0(1 x^22+x^48 ...) (1 x^22+x^44! ...)x^4 =18 ...

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Standard XII

Mathematics

Expansions to Remove Indeterminate Form

Let $ fx= beg...

Question

Let $f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{e^{\frac{- x^{2}}{2}} - cos x}{x ln (1 + x) sin x (e^{x} - 1)}, & x \neq 0 k, & x = 0 \end{matrix} .$
If $f (x)$ is continuous at $x = 0$ , then $k$ equals

\frac{1}{4}

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\frac{1}{6}

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\frac{1}{12}

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\frac{1}{8}

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Solution

The correct option is C $\frac{1}{12}$
$k = lim x \to 0 \frac{e^{\frac{- x^{2}}{2}} - cos x}{x ln (1 + x) sin x (e^{x} - 1)}$

$k = lim x \to 0 \frac{e^{\frac{- x^{2}}{2}} - cos x}{x^{4}}$

$= lim x \to 0 \frac{(1 - \frac{x^{2}}{2} + \frac{x^{4}}{8} - . . .) - (1 - \frac{x^{2}}{2} + \frac{x^{4}}{4!} - . . .)}{x^{4}}$

$= \frac{1}{8} - \frac{1}{24} = \frac{1}{12}$

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Let f(x)=⎧⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪⎩e−x22−cosxxln(1+x)sinx(ex−1), x≠0 k , x=0. If f(x) is continuous at x=0, then k equals

The correct option is C 112k=limx → 0e−x22−cosxxln(1+x)sinx(ex−1) k=limx → 0e−x22−cosxx4 =limx → 0(1−x22+x48−...)−(1−x22+x44!−...)x4 =18−124=112

Let $f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{e^{\frac{- x^{2}}{2}} - cos x}{x ln (1 + x) sin x (e^{x} - 1)}, & x \neq 0 k, & x = 0 \end{matrix} .$
If $f (x)$ is continuous at $x = 0$ , then $k$ equals