If f x - begin cases xsin frac 1x- x neq 0 1 0 - x -0lend cases - then lim x - 0 fx-A- 0B- -1C- 1D- does not exist

The correct option is A 0 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 sandwic ...

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Standard XI

Mathematics

Sandwich Theorem

If f x = begi...

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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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 $l i m_{x \to 0} f (x) = 0$