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I'm currently struggling with how to solve the following equation without L'Hopital's rule. Everywhere I've found online only uses L'Hopital's method.

$$\lim_{x \to \pi/2}\left({\sin x -1\over \cos x}\right)$$

Any help is appreciated!

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closed as off-topic by Namaste, user296602, Adrian Keister, Theoretical Economist, Rafa Budría Sep 22 '18 at 4:28

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Multiply numerator and denominator by $\sin x+1$: you get $$ \lim_{x\to \pi/2}\frac{-\cos^2x}{\cos x(\sin x+1)} $$

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  • $\begingroup$ Where would I go from here now? Would I divide the $\cos x$ out of the denominator giving me: $$\lim_{x \to \pi/2}\left({-\cos x \over \sin x +1}\right)$$ Then continue on to substitute giving me: $ 0 \over 2$ or 0? $\endgroup$ – Kal Sep 21 '18 at 22:32
  • $\begingroup$ @Kal Exactly so. $\endgroup$ – egreg Sep 21 '18 at 22:33
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Here is an alternative approach different from egreg's contribution. Remember that: \begin{align*} \sin(x) - 1 = 2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right) - \sin^{2}\left(\frac{x}{2}\right) - \cos^{2}\left(\frac{x}{2}\right) = - \left(\sin\left(\frac{x}{2}\right) - \cos\left(\frac{x}{2}\right)\right)^{2} \end{align*} Analogously, we have: \begin{align*} \cos(x) = \cos^{2}\left(\frac{x}{2}\right) - \sin^{2}\left(\frac{x}{2}\right) = \left[\cos\left(\frac{x}{2}\right) - \sin\left(\frac{x}{2}\right)\right]\left[\cos\left(\frac{x}{2}\right) + \sin\left(\frac{x}{2}\right)\right] \end{align*} Therefore the sought limit is given by: \begin{align*} \lim_{x\rightarrow\pi/2}\frac{\sin(x) - 1}{\cos(x)} = \lim_{y\rightarrow\pi/4}\frac{\sin(y)-\cos(y)}{\cos(y)+\sin(y)} = 0 \end{align*}

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A lot of trig. identities in the given answers. But that's a lot of buckshot scattered around the barn. Suppose we have $f'(a) = 0$ and $g(a)=0,$ $g'(a)\ne 0.$ Then $g(x)\ne 0$ for $x$ near and not equal to $a,$ and

$$\frac{f(x)-f(a)}{g(x)} = \frac{f(x)-f(a)}{g(x)-g(a)} = \frac{(f(x)-f(a))/(x-a)}{(g(x)-g(a))/(x-a)}$$ $$ \to f'(a)/g'(a) = \frac{0}{g'(a)}=0.$$

In the given problem $f(x) = \sin x, g(x)=\cos x, a=\pi/2.$

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We have by $x=\pi/2 -y$ with $y \to 0$

$$\lim_{x \to \pi/2}\left({\sin x -1\over \cos x}\right)=\lim_{y \to 0}\left({\cos y -1\over \sin y}\right)=0$$

indeed by standard limits

$$\frac{\cos y -1}{ \sin y}=\frac{\cos y -1}{ y^2}\,\frac{y}{ \sin y}\,y\to -\frac12\cdot 1\cdot 0=0$$

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Recall the well known angle sum identities https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Angle_sum_and_difference_identities

Let $l$ denote our limit and suppose that the limit exists. We have

$$l = lim_{y\rightarrow0}\frac{\sin{(\pi / 2 + y)} - 1}{cos{(\pi /2 +y)}} = lim_{y\rightarrow0}\frac{\cos{(y)} - 1}{-sin{(y)}}$$ but $$l = lim_{z\rightarrow0}\frac{\sin{(\pi / 2 - z)} - 1}{cos{(\pi /2 -z)}} = lim_{z\rightarrow0}\frac{\cos{(z)} - 1}{sin{(z)}}$$ Thus $l = -l$

Thus $l = 0$

Now, suppose that the limit doesn't exist. Then we have $\frac{\cos{(y)} - 1}{-sin{(y)}} = tan(y/2)$ so that $lim_{y \rightarrow 0}\tan{(\frac y2)}$ doesn't exist. A contradiction since $\tan{\frac y2}$ is continuous about zero.

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  • $\begingroup$ This only shows that if the limit exists then it is $0$. But why does it exist to begin with? $\endgroup$ – egreg Sep 23 '18 at 9:04
  • $\begingroup$ I thought the question presupposed the existence of the limit. "How to solve the limit" translates in my mind to "what must the limit, which we know to exist, be?" $\endgroup$ – Sid Sep 25 '18 at 15:46
  • $\begingroup$ It has been fixed. $\endgroup$ – Sid Sep 25 '18 at 19:33

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