## Calculate the limit using geometric mean

Home -> Solved problems -> Limit geometric mean ### Solution

In the next figure we have a circle arc of radius 1 and center $$A$$. Let $$S$$ be the area of the triangle $$\triangle ABC$$ thus, $S=\frac{1}{2}AB\sin x=\frac{1}{2}\sin x$ Let $$W$$ be the area of the circular sector shaded in grey in the next figure, $W=\frac{1}{2}x$ $$T$$ is the area of the triangle $$\triangle ABD$$$T=\frac{1}{2}ABDB=\frac{1}{2}\tan x$ By inclusion, we have $$T\geq W\geq S$$ thus, $\frac{1}{2}\tan x\geq \frac{1}{2}x\geq \frac{1}{2}\sin x$ Dividing the previous inequality by $$\frac{1}{2}\sin x$$ and taking reciprocals, we get $\cos x\leq\frac{\sin x}{x}\leq1$ Since $$x\rightarrow\frac{\sin x}{x}$$ and $$x\rightarrow\cos x$$ are two even functions the inequality $$\cos x\leq\frac{\sin x}{x}\leq1$$ is valid for any non-zero $$x$$ between $$-\frac{\pi}{2}$$ and $$-\frac{\pi}{2}$$. Finally, knowing that $$\cos0=1$$ and using squeeze theorem we get that $\lim_{x \rightarrow 0}\frac{\sin x}{x}=1$
Home -> Solved problems -> Limit geometric mean

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