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I don't know how to solve the following:

$$\lim_{u \to 2} \frac{1}{(2-u) \left( \sqrt{\frac{u+2}{u-1}}-2\right)}$$

and

$$\lim_{x \to 2} \frac{ \sin(1-\cos x)}{x \tan(\pi x)} $$

I think I may need to multiply the second by $1-\cos x$, but I am not sure.

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  • $\begingroup$ Are you asking $\lim_{u\to2}{\frac{1}{2-u(\sqrt\frac{u+2}{u-1}-2)}}$ and $\lim_{x\to2}{\frac{\sin(1-\cos x)}{x\tan(\pi x)}}$? $\endgroup$ – Marconius Oct 21 '15 at 0:58
  • $\begingroup$ @SUZUKI I edited making the best guess as to what functions you intended. Please edit the expressions if they're wrong. This should be pretty intuitive when you see the latex code, but feel free to let me know the changes needed if you can't make sense of it. $\endgroup$ – stochasticboy321 Oct 21 '15 at 0:59
  • $\begingroup$ The second limit should be with $x \to 0$ otherwise the limit does not exist. $\endgroup$ – Paramanand Singh Oct 21 '15 at 4:00
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In both cases no two sided limit exists. For the first function, multiply by the conjugate twice \begin{align} \lim_{u \to 2} \frac{1}{(2 - u)(\sqrt{\frac{u + 2}{u - 1}} - 2)} &= \frac{1}{(2 - u)(\sqrt{\frac{u + 2}{u - 1}} - 2)} \cdot \frac{(2 + u)(\sqrt{\frac{u + 2}{u - 1}} + 2)}{(2 + u)(\sqrt{\frac{u + 2}{u - 1}} + 2)}\\ &= \frac{(u + 2)(4)}{(u^2 - 4)(\frac{u + 2}{u - 1} - 4)}\\ &= \frac{(u + 2)(4)}{(u^2 - 4)(\frac{1}{u - 1} \cdot) -3(u + 2)}\\ &= \frac{4(u - 1)}{-3(u^2 - 4)}\\ &= \frac{-4}{3}(\frac{\frac{1}{2}}{u - 2} + \frac{\frac{1}{2}}{u + 2}) = \frac{-4}{3}\lim_{u \to 2}(\frac{1}{2(u - 2)} + \frac{1}{8}) \end{align} And since the general reciprocal function has no two sided limit at $x \to 0$, this function has no two sided limit for $u \to 2$. And for the second function, the top is continuos meaning it has the same limit has $\frac{1}{x\tan(\pi x)}$ which has no two sided limit.

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