Question

this question of limits chapter​

Answer 1

Answer:

lim

x

→

0

x

2

sin

(

1

x

)

Since

x

2

⋅

−

1

≤

x

2

sin

(

1

x

)

≤

x

2

⋅

1

and

lim

x

→

0

x

2

⋅

−

1

=

lim

x

→

0

x

2

⋅

1

=

0

, apply the squeeze theorem.

0

Answer 2

Answer:

$lt_{x - > 0}( {x}^{2} \sin( \frac{1}{x} ) ) = 0 \times \sin( \infty ) \\ \\ = 0 \times (value \: \: between \: \: ( - 1 \: . \: 1)) \\ \\ = 0$