If fx-tan x-x- x- 0 1 - x-0 -The function at x-0 is

The correct options are A differentiable B continuous nbsp;lim_x → 0^+ nbsp;f(x)=tanh/h=1 lim_x → 0^ f(x)= tanh/ h=1 Hene LHL = RHL= f(0) So it ...

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Relative Velocity

If fx=tan x...

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If $f (x) = {\begin{matrix} \frac{tan x}{x}, & x \neq 0 1, & x = 0 \end{matrix}$ .The function at $x = 0$ is

differentiable

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continuous

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discontinuous

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left continuous only

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f (x) = \{\begin{matrix} \frac{\sin^{2} a x}{x^{2}}, & x \neq 0 \\ 1, & x = 0 \end{matrix}

f (x) = {\begin{matrix} \frac{tan x}{x}, x \neq 0 1, x = 0 \end{matrix}

f (x)

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f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \begin{matrix} {(1 - | tan x |)}^{a / | tan x |}, & - \frac{π}{4} < x < 0 \end{matrix} \begin{matrix} b, & x = 0 \end{matrix} \begin{matrix} e^{\frac{sin 3 x}{sin 2 x}}, & 0 < x < \frac{π}{4} \end{matrix} \end{matrix}

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f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} (1 + | tan x |)^{\frac{p}{| tan x |}} & , - \frac{π}{3} < x < 0 q & x = 0 e^{\frac{sin 3 x}{sin 2 x}}, & 0 < x < \frac{π}{3} \end{matrix}

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f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} \frac{{sin}^{2} a x}{x^{2}}, x \neq 0 1, x = 0 \end{matrix}

x = 0

a =

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