For each t [ ] R , let [t] be the greatest integer less than or equal to t Then [ _x 1^ + …

Question

For each t $$ \in $$ R , let [t] be the greatest integer less than or equal to t

Then  $$\mathop {\lim }\limits_{x \to 1^ + } {{\left( {1 - \left| x \right| + \sin \left| {1 - x} \right|} \right)\sin \left( {{\pi  \over 2}\left[ {1 - x} \right]} \right)} \over {\left| {1 - x} \right|.\left[ {1 - x} \right]}}$$

Accepted Answer

$$\mathop {\lim }\limits_{x \to {1^ + }} {{\left( {1 - \left| x \right| + \sin \left| {1 - x} \right|} \right)\sin \left( {{\pi  \over 2}\left[ {1 - x} \right]} \right)} \over {\left| {1 - x} \right|\left[ {1 - x} \right]}}$$

$=$ $$\mathop {\lim }\limits_{x \to {1^ + }} {{\left( {1 - x} \right) + \sin \left( {x - 1} \right)} \over {\left( {x - 1} \right)\left( { - 1} \right)}}$$ $$\sin \left( {{\pi  \over 2}\left( { - 1} \right)} \right)$$

$=$ $$\mathop {\lim }\limits_{x \to {1^ + }} \left( {1 - {{\sin \left( {x - 1} \right)} \over {\left( {x - 1} \right)}}} \right)\left( { - 1} \right) = \left( {1 - 1} \right)\left( { - 1} \right) = 0$$

Solutions · 1