What a Limit!

Calculus Level 5

lim ⁡ x → 0 ⌊ ( 1 − e x ) sin ⁡ x ∣ x ∣ ⌋ = ? \large \lim_{x\rightarrow 0}\left\lfloor (1-e^{x})\frac{\sin x}{\left | x \right |} \right \rfloor = \, ?

Notations:

0 Limit does not exist -1 1

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1 solution

Consider the limits when x → 0 + x \to 0^+ and x → 0 − x \to 0^- separately.

L + = lim ⁡ x → 0 + ⌊ ( 1 − e x ) sin ⁡ x ∣ x ∣ ⌋ = lim ⁡ x → 0 + ⌊ ( 1 − e x ) sin ⁡ x x ⌋ = ⌊ ( 1 − e 0 + ) sin ⁡ 0 + 0 + ⌋ = ⌊ ( 1 − e 0 + ) 1 ⌋ Note that e 0 + > 1 = − 1 \begin{aligned} L_+ & = \lim_{x \to 0^+} \left \lfloor (1-e^x) \frac {\sin x}{\color{#3D99F6}|x|} \right \rfloor \\ & = \lim_{x \to 0^+} \left \lfloor (1-e^x) \frac {\sin x}{\color{#3D99F6}x} \right \rfloor \\ & = \left \lfloor (1-e^{0^+}) {\color{#3D99F6}\frac {\sin 0^+}{0^+}} \right \rfloor \\ & = \left \lfloor (1-{\color{#D61F06}e^{0^+}}) {\color{#3D99F6}1} \right \rfloor & \small \color{#D61F06} \text{Note that }e^{0^+} > 1 \\ & = - 1 \end{aligned}

L − = lim ⁡ x → 0 − ⌊ ( 1 − e x ) sin ⁡ x ∣ x ∣ ⌋ = lim ⁡ x → 0 − ⌊ ( 1 − e x ) sin ⁡ x − x ⌋ = ⌊ ( 1 − e 0 − ) sin ⁡ 0 − − 0 − ⌋ = ⌊ ( 1 − e 0 − ) ( − 1 ) ⌋ Note that e 0 + < 1 = − 1 \begin{aligned} L_- & = \lim_{x \to 0^-} \left \lfloor (1-e^x) \frac {\sin x}{\color{#D61F06}|x|} \right \rfloor \\ & = \lim_{x \to 0^-} \left \lfloor (1-e^x) \frac {\sin x}{\color{#D61F06}-x} \right \rfloor \\ & = \left \lfloor (1-e^{0^-}) {\color{#D61F06}\frac {\sin 0^-}{-0^-}} \right \rfloor \\ & = \left \lfloor (1-{\color{#3D99F6}e^{0^-}}) ({\color{#D61F06}-1}) \right \rfloor & \small \color{#3D99F6} \text{Note that }e^{0^+} < 1 \\ & = - 1 \end{aligned}

Since L + = L − = − 1 ⟹ L = − 1 L_+ = L_- = -1\implies L = \boxed{-1}

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( link try to do this problem please

Sudhamsh Suraj - 4 years, 3 months ago

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