Suppose x ≠ 0. x\not=0. x = 0.
0 ≤ ∣ sin ( e x ) x ∣ ≤ 1 ∣ x ∣ , x ∈ R , x ≠ 0 0\leq|\dfrac{\sin(e^x)}{x}|\leq \dfrac{1}{|x|}, x\in\R, x\not=0 0 ≤ ∣ x sin ( e x ) ∣ ≤ ∣ x ∣ 1 , x ∈ R , x = 0
lim x → ∞ 1 ∣ x ∣ = 0 \lim\limits_{x\to\infin}\dfrac{1}{|x|}=0 x → ∞ lim ∣ x ∣ 1 = 0 Then by the Squeeze Theorem
lim x → ∞ ∣ sin ( e x ) x ∣ = 0 \lim\limits_{x\to\infin}|\dfrac{\sin(e^x)}{x}|=0 x → ∞ lim ∣ x sin ( e x ) ∣ = 0
0 ≤ ∣ sin ( e x ) x 2 + 2 ∣ ≤ ∣ sin ( e x ) x ∣ , x ∈ R , x ≠ 0 0\leq|\dfrac{\sin(e^x)}{\sqrt{x^2+2}}|\leq |\dfrac{\sin(e^x)}{x}|, x\in\R, x\not=0 0 ≤ ∣ x 2 + 2 sin ( e x ) ∣ ≤ ∣ x sin ( e x ) ∣ , x ∈ R , x = 0
Then by the Squeeze Theorem
lim x → ∞ ∣ sin ( e x ) x 2 + 2 ∣ = 0 \lim\limits_{x\to\infin}|\dfrac{\sin(e^x)}{\sqrt{x^2+2}}|=0 x → ∞ lim ∣ x 2 + 2 sin ( e x ) ∣ = 0 Therefore
lim x → ∞ sin ( e x ) x 2 + 2 = 0 \lim\limits_{x\to\infin}\dfrac{\sin(e^x)}{\sqrt{x^2+2}}=0 x → ∞ lim x 2 + 2 sin ( e x ) = 0
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