Iffx-xsin 1-x x - 00 x-0-then limx- 0 fx-

Here f(0)=0 Since 1 ≤ sin1/x ≤ 1 ⇒ |x| ≤ xsin1/x ≤ |x| We know that lim_x→ 0 |x| = 0 and lim_x→ 0 |x|=0 With the help of sandwich theorem we can say lim_ ...

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Byju's Answer

Standard XIII

Mathematics

Sandwich Theorem

Iffx=xsin 1/x...

Question

If

$f (x) = {\begin{matrix} x s i n \frac{1}{x} & x \neq 0 0 & x = 0 \end{matrix},$

then $lim x \to 0 f (x) =$

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-1

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does not exist

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Solution

The correct option is B
0

Here $f (0) = 0$
Since $- 1 \leq sin \frac{1}{x} \leq 1 \Rightarrow - | x | \leq x sin \frac{1}{x} \leq | x |$

We know that $l i m_{x \to 0} | x | = 0$ and $l i m_{x \to 0} - | x | = 0$

With the help of sandwich theorem we can say $l i m_{x \to 0} f (x) = 0$

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Similar questions

$f (x) = {\begin{matrix} x s i n \frac{1}{x} & x \neq 0 0 & x = 0 \end{matrix},$

then $lim x \to 0 f (x) =$

Q. If

f (x) = x sin (1 / x), x \neq 0

, then

{lim}_{x \to 0} f (x) =

Q. If

f (x) = x \sin (1 / x), x \neq 0,

then

\lim_{x \to 0} f (x) =

(a) 1
(b) 0
(c) −1
(d) does not exist

If $f (x) = x s i n (\frac{1}{x}), x \neq 0,$ then ${lim}_{x \to 0} f (x) =$

Q. Evaluate

\lim_{x \to 0} f (x)

, where

f (x) = \{\begin{cases} \frac{|x|}{x}, & x \neq 0 \\ 0, & x = 0 \end{cases}

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If f(x)={xsin1xx≠00x=0, then limx→0f(x)=

The correct option is B 0 Here f(0)=0 Since −1≤sin1x≤1⇒−|x|≤xsin1x≤|x| We know that limx→0|x|=0 and limx→0−|x|=0 With the help of sandwich theorem we can say limx→0f(x)=0

If

$f (x) = {\begin{matrix} x s i n \frac{1}{x} & x \neq 0 0 & x = 0 \end{matrix},$

then $lim x \to 0 f (x) =$

The correct option is B
0

Here $f (0) = 0$
Since $- 1 \leq sin \frac{1}{x} \leq 1 \Rightarrow - | x | \leq x sin \frac{1}{x} \leq | x |$

We know that $l i m_{x \to 0} | x | = 0$ and $l i m_{x \to 0} - | x | = 0$

With the help of sandwich theorem we can say $l i m_{x \to 0} f (x) = 0$