Suppose we know that $- x ^ { 2 } \leq x ^ { 2 } \sin \left( \frac { \pi } { x } \right) \leq x ^ { 2 } \text { for } x \in [ - 1,1 ]$ Use the Squeeze Theorem to determine $\lim _ { x \rightarrow 0 } x ^ { 2 } \sin \left( \frac { \pi } { x } \right)$

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The Answer of Suppose we know that $- x ^ { 2 } \leq x ^ { 2 } \sin \left( \frac { \pi } { x } \right) \leq x ^ { 2 } \text { for } x \in [ - 1,1 ]$ Use the Squeeze Theorem to determine $\lim _ { x \rightarrow 0 } x ^ { 2 } \sin \left( \frac { \pi } { x } \right)$

Suppose We Know That −x2≤x2sin⁡(πx)≤x2 for x∈[−1,1]- x ^ { 2 } \leq x ^ { 2 } \sin \left( \frac { \pi } { x } \right) \leq x ^ { 2 } \text { for } x \in [ - 1,1 ]−x2≤x2sin(xπ​)≤x2 for x∈[−1,1]

Suppose We Know That $- x ^ { 2 } \leq x ^ { 2 } \sin \left( \frac { \pi } { x } \right) \leq x ^ { 2 } \text { for } x \in [ - 1,1 ]$