lim-x-senx-x-

Question Number 62180 by Mikael last updated on 17/Jun/19

$$\underset{{x}\rightarrow\infty} {{lim}}\:\frac{{senx}}{{x}} \\ $$

Commented by maxmathsup by imad last updated on 17/Jun/19

if you mean sinx we have ∣sinx∣≤1 ⇒∣((sinx)/x)∣≤(1/(∣x∣)) for all x≠0 but lim_(x→∞) (1/(∣x∣)) =0 ⇒ lim_(x→∞) ((sinx)/x) =0

$${if}\:{you}\:{mean}\:{sinx}\:\:{we}\:{have}\:\:\mid{sinx}\mid\leqslant\mathrm{1}\:\Rightarrow\mid\frac{{sinx}}{{x}}\mid\leqslant\frac{\mathrm{1}}{\mid{x}\mid}\:\:{for}\:{all}\:{x}\neq\mathrm{0}\:\:{but} \\ $$$${lim}_{{x}\rightarrow\infty} \:\:\frac{\mathrm{1}}{\mid{x}\mid}\:=\mathrm{0}\:\Rightarrow\:{lim}_{{x}\rightarrow\infty} \:\frac{{sinx}}{{x}}\:=\mathrm{0}\:\:\: \\ $$

Commented by Tony Lin last updated on 17/Jun/19

∀ε>0⇒0≤∣((senx)/x)∣≤∣(1/x)∣∀x≥(1/ε) ∵lim_(x→∞) 0=0 and lim_(x→∞) (1/x)=0 ⇒lim_(x→∞) ((senx)/x)=0(squeeze theoreom)

$$\forall\varepsilon>\mathrm{0}\Rightarrow\mathrm{0}\leqslant\mid\frac{{senx}}{{x}}\mid\leqslant\mid\frac{\mathrm{1}}{{x}}\mid\forall{x}\geqslant\frac{\mathrm{1}}{\varepsilon} \\ $$$$\because\underset{{x}\rightarrow\infty} {\mathrm{lim}0}=\mathrm{0}\:{and}\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{x}}=\mathrm{0} \\ $$$$\Rightarrow\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{{senx}}{{x}}=\mathrm{0}\left({squeeze}\:{theoreom}\right) \\ $$

Commented by Mikael last updated on 17/Jun/19