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Using sequential criterion for functional limits, show that $$\lim_{x\to \frac{\pi}{2}}\tan{x}$$ does not exists.

Let $f(x)=\tan{x}$. The domain of $f(x)$ is $S=\mathbb{R}-\{(2n+1)\frac{\pi}{2}\}$, $n\in \mathbb{N}$.

I have to chosse two sequences $\{x_n\}$ and $\{y_n\}$ which converge to same limit but $\lim f(x_n)\neq \lim f(y_n)$.

Please suggest two sequences $\{x_n\}$ and $\{y_n\}$ within the domain.

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Pick $x_n$ an increasing sequence converging to $\frac{\pi}{2}$ and $y_n$ a decreasing sequence converging to $\frac{\pi}{2}$. Note that $\tan(x_n)>0$ but $\tan(y_n)<0$ (look at the signs of sine and cosine).

So the only possibility now is that $\lim_{x\to \frac\pi2} \tan x=0$ (why?), which cannot happen, because then

$$\lim_{x\to \frac\pi2} \sin x=\lim_{x\to \frac\pi2}\cos x\cdot \tan x=\lim_{x\to \frac\pi2}\cos x\cdot \lim_{x\to \frac\pi2}\tan x=0$$

and $\lim_{x\to \frac\pi2} \sin x=1$.

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  • $\begingroup$ Your point in clear to me. But which $x_n$ and $y_n$ should I choose is still not identified. $\endgroup$ – user1942348 Feb 27 '17 at 13:31
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    $\begingroup$ Choose whichever: for example $y_n=\frac\pi2 + \frac1n$, $x_n=\frac\pi2 - \frac1n$ $\endgroup$ – A. Salguero-Alarcón Feb 27 '17 at 13:32

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