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Existence of Limit

If f x =| x...

Question

If $f (x) = {| x |}^{sin x}$ . then

A

lim x \to 0^{-} f (x) = 1

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B

lim x \to 0^{+} f (x) = 1

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C

lim x \to 0 f (x) = 1

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D

Limit does not exist at x-0

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Solution

The correct options are
A $lim x \to 0^{-} f (x) = 1$
B $lim x \to 0^{+} f (x) = 1$
C $lim x \to 0 f (x) = 1$

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Similar questions

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{tan x - sin x}{x^{3}}; & x < 0 \frac{{cot}^{- 1} x - {cos}^{- 1} x}{x^{3}}; & x > 0 \frac{1}{2}; & x = 0 \end{matrix}

Then which of the following is correct

Let f (x) = \frac{1 - x (1 + | 1 - x |)}{| 1 - x |} cos (\frac{1}{1 - x})

x \neq 1.

f (x) = \frac{x + | x | (1 + x)}{x} sin (\frac{1}{x}), x \neq 0

L = lim x \to 0^{-} f (x)

R = lim x \to 0^{+} f (x) .

f (x) ⎧ ⎨ ⎩ \begin{matrix} m x^{2} + n, x < 0 n x + m, 0 \leq x \leq n x^{3} + m, x > 1 \end{matrix} 1

For what integers

m

n

lim x \to 0 f (x)

lim x \to 1 f (x)

Q. Assertion :If

lim x \to 0 f (x)

lim x \to 0 g (x)

exists finitely, then

lim x \to 0 f (x) \cdot g (x)

exists finitely. Reason: If

lim x \to 0 f (x) \cdot g (x)

exists finitely then

lim x \to 0 f (x) \cdot g (x) = lim x \to 0 f (x) \cdot lim x \to 0 g (x)

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