# L'Hopital's rule

$$\lim_{x\to \pi/2} (2 x - \pi) \sec(x) = 0 \times \infty$$

can be solved by putting it on the form:

$$\lim_{x\to \pi/2} (2 x - \pi) / (1/sec(x))$$

and by using L'Hopital's rule. The answer will be $-2$.

"My question is can we also solving it using" L'Hopital's rule " after putting it on the form:

$$\lim_{x\to \pi/2} \sec(x)/(1/(2 x - \pi))$$

• Every form $\frac{\infty}{\infty}$ ,$\frac{0}{0}$ , $0\cdot\mp\infty$ , etc can be solved using L'Hospital's rule IF $x\in\mathbb{R}$. – Theodoros Mpalis Mar 29 '16 at 4:17

## 3 Answers

write it in the form $$\lim_{x \to \frac{\pi}{2}}\frac{2x-\pi}{\cos(x)}=\lim_{x \to \frac{\pi}{2}}\frac{2}{-\sin(x)}=\frac{2}{-1}=-2$$

• It's clearer if you add $= \frac{2}{\sin(x)}$ after the first equality. – marty cohen Mar 29 '16 at 5:03
• @Dr.Sonnhard Graubner Sorry I correct my question please check it again – Taha Topology Mar 29 '16 at 5:07

Well, you can solve it with L'Hôspital rule as well :$$\lim_{x\rightarrow\frac{\pi}{2}}(2x-\pi)\sec(x)=\lim_{x\rightarrow\frac{\pi}{2}}\frac{2x-\pi}{{1}\over{\sec(x)}}$$ $$=\lim_{x\rightarrow\frac{\pi}{2}}\frac{2}{{\tan(x)}\over{\sec^2(x)\cos(x)}}$$ $$=\lim_{x\rightarrow\frac{\pi}{2}}\frac{2\sec^2(x)\cos(x)}{\tan(x)}$$ $$=\lim_{x\rightarrow\frac{\pi}{2}}\frac{{2}\over{\cos(x)}}{\tan(x)}=\lim_{x\rightarrow\frac{\pi}{2}}\frac{2}{\tan(x)\cos(x)}.$$ Now, because of $$\lim_{x\rightarrow\frac{\pi}{2}}\tan(x)\cos(x)=-1\Longleftrightarrow\lim_{x\rightarrow\frac{\pi}{2}}\frac{1}{\tan(x)\cos(x)}=\frac{1}{-1}=-1.$$ Now simplify: $$\lim_{x\rightarrow\frac{\pi}{2}}\frac{2}{\tan(x)\cos(x)}=\frac{2}{-1}=-2.$$

• appreciate your cooperation with me greatly....But my point is solving it starting from the form limit sec(x)/(1/(2*x-pi); and using D.L.H – Taha Topology Mar 29 '16 at 6:24
• @TahaTopology It cannot be solved as sec(x)/(1/2x–π). – Theodoros Mpalis Mar 29 '16 at 6:26
• @TahaTopology Some times on limits $$\frac{f(x)}{{1}\over{g(x)}}\neq\frac{g(x)}{{1}\over{f(x)}}.$$ – Theodoros Mpalis Mar 29 '16 at 6:29
• Then I did huge mistake – Taha Topology Mar 29 '16 at 6:30
• @TahaTopology This one can only be solved if you transform it to the form: $$\frac{2x-\pi}{{1}\over{\sec(x)}}$$ and cannot be solved as: $$\frac{\sec(x)}{{1}\over{2x-\pi}}.$$ – Theodoros Mpalis Mar 29 '16 at 6:34

No, because $$\lim_{x\rightarrow \frac{\pi}{2}}(2x-\pi)\sec(x)\neq\lim_{x\rightarrow\frac{\pi}{2}}\frac{2x-\pi}{\sec(x)}.$$ But equal to $$\lim_{x\rightarrow\frac{\pi}{2}}\frac{\sec(x)}{{1}\over{2x-\pi}}.$$

• Mpalis Yes thank you I correct my question ,please check it again . – Taha Topology Mar 29 '16 at 5:09
• Yes you can, why not ? – Theodoros Mpalis Mar 29 '16 at 5:11
• @TahaTopology As I said, all these forms can be solved with L'Hôspital – Theodoros Mpalis Mar 29 '16 at 5:12
• using L'Hopital's rule? please can you show me how ??? when I get the derivative of sec(x)=sec(x) *tan(x) ....and 1/(2*x-pi) it will get -2/(2*x-pi)^2 .....it getting complicated ? – Taha Topology Mar 29 '16 at 5:16
• @Theodors Mpalis ...ok lim tan(x)/((cos(x)/-2*(2*x-pi)^(2))=inifinty/(0/infinity) ???? for me it's very hard to get (-2) ..please more explane – Taha Topology Mar 29 '16 at 5:25