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Prove that $$\lim_{x\to \frac{\pi}{2}} \frac{1}{\big(x-\frac{\pi}{2}\big)}+{\tan(x)}=0.$$

I'm not really sure how to proceed. I know that I should not try L'Hôpital's rule (tried that) but not sure how I would incorporate into the Squeeze Theorem or how I would use continuity.

Thanks!

Edit: Turns out I was really dumb and you do use L'Hôpital's rule twice. I made the mistake of differentiating the whole quotient rather than the function on top and the bottom of the vinculum separately.

k170
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J. Cricks
  • 515

4 Answers4

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Let $t=x-\frac{\pi}{2}$, then we have to compute

$$\lim_{t \to 0}\frac{1}{t}-\cot t.$$ Now use the series expansion for $\cot t$ given by $$\cot t=\frac{\cos t}{\sin t}=\left(1-\frac{t^2}{2}+o(t^3)\right)\frac{1}{t}\left(1+\frac{t^2}{6}+o(t^3)\right)=\frac{1}{t}-\frac{t}{3}+o(t).$$ So $$\lim_{t \to 0}\frac{1}{t}-\cot t=\lim_{t \to 0}\frac{t}{3}+o(t)=0.$$

Anurag A
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Sub $u=\frac{\pi}{2}-x$ then we have

$$ \begin{align} &\lim_{u\to0} \frac{1}{u}-\cot u \\ &=\lim_{u\to0} \frac{1-u\cot u}{u} \\ &=\lim_{u\to0} \left(-\cot u+u\csc^2u \right) \\ &=\lim_{u\to0} \left(\frac{u}{\sin^2u}-\frac{\cos u}{\sin u} \right) \\ &=\lim_{u\to0} \frac{u-\sin u\cos u}{\sin^2u} \\ &=\lim_{u\to0} \frac{u-\sin u\cos u}{2\sin u\cos u} \\ &=\lim_{u\to0} \left(\frac{u}{2\sin u\cos u} - \frac{1}{2} \right) \\ &=0 \end{align} $$

bFur4list
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Let $y= \dfrac{\pi}{2}-x$

And rewrite as

$\lim_y \rightarrow 0 \dfrac{\tan y - y}{y^2} \dfrac{y}{\tan y}$

Hari Shankar
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First note that $$\lim_{x\to0}\frac{\sin(x)}{x}=1$$ $$\lim_{x\to0}\frac{x}{\sin(x)}=1$$ $$\lim_{x\to0}\frac{1-\cos(x)}{x}=0$$ Keeping these fundamental trigonometric limits in mind, we have $$\lim_{x\to\frac{\pi}{2}} \frac{1}{\left(x-\frac{\pi}{2}\right)}+\tan(x)$$ $$=\lim_{t\to0} \frac{1}{t}-\cot(t)$$ $$=\lim_{t\to0}\frac{\sin(t)-t\cos(t)}{t\sin(t)}$$ $$=\left(\lim_{t\to0}\frac{\sin(t)}{t}\right)\left(\lim_{t\to0}\frac{\sin(t)-t\cos(t)}{\sin^2(t)}\right)$$ $$=\left(\lim_{t\to0}\frac{\sin(t)}{t}\right)^2\left(\lim_{t\to0}\frac{t-t\cos(t)}{\sin^2(t)}\right)$$ $$=\left(\lim_{t\to0}\frac{\sin(t)}{t}\right)^2\left(\lim_{t\to0}\frac{t}{\sin(t)}\right)^2\left(\lim_{t\to0}\frac{1-\cos(t)}{t}\right)$$ Let me know if this helps you.

k170
  • 9,045