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Find $$\lim_{x\to\pi/4} \frac{1-\tan(x)^2}{\sqrt{2}\times \cos(x)-1}$$ without using L'Hôpital's rule.

I can solve it using L'Hôpital's rule, but is it possible to solve it without using L'Hôpital's rule?

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  • $\begingroup$ You want to use the (limit) definition of derivative for $\tan$ and $\cos$ at $\frac{\pi}{4}$. $\endgroup$ – Matthew Leingang Dec 29 '16 at 1:47
  • $\begingroup$ OP asked for "without l'Hospital's rule". $\endgroup$ – Oscar Lanzi Dec 29 '16 at 2:57
  • $\begingroup$ Shouldn't that be written $\tan^2 x?$ $\endgroup$ – zhw. Dec 29 '16 at 4:10
  • $\begingroup$ @OscarLanzi The definition of the derivative is not L'Hopital. $\endgroup$ – zhw. Dec 29 '16 at 4:11
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You can try the following: \begin{align} \frac{1-\tan^2({x})}{\sqrt{2} \cos{x}-1}\frac{\sqrt{2} \cos{x}+1}{ \sqrt{2} \cos{x}+1 }&=\frac{1-\frac{\sin^2{x}}{\cos^2{x}}}{2 \cos^2{x}-1} (\sqrt{2} \cos{x}+1)\\ &=\frac{\cos^2{x}-\sin^2{x}}{\cos^2{x} (2\cos^2{x}-\sin^2{x}-\cos^2{x}) }(\sqrt{2} \cos{x}+1)\\ &=\frac{\sqrt{2} \cos{x}+1}{\cos^2x}. \end{align} Now take the limit as usual.

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Let $f(x) = \tan^2 x, g(x) = \sqrt 2 \cos x.$ The expression equals

$$-\frac{f(x) - f(\pi/4)}{g(x) - g(\pi/4)} = -\frac{(f(x) - f(\pi/4))/(x-\pi/4)}{(g(x) - g(\pi/4))/(x-\pi/4)} .$$

As $x\to \pi/4,$ the last expression $\to -f'(\pi/4)/g'(\pi/4)$ by definition of the derivative, which is a simple computation. (No L'Hopital was used here.)

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Change the variable to $y=x-\frac{\pi}{2}$. then the limit is

$$\frac{-4\sin y \cos y}{(\cos y-\sin y)^2 (\cos y-\sin y -1)}$$

Now both $\cos y$ and $(\cos y-\sin y)^2$ have a limit of $1$ so we are reduced to $$\frac{-4\sin y}{\cos y-\sin y -1}$$

Now $\frac{\sin y}{y}\to 1$ so we need to find the limit of $$\frac{-4 y}{\cos y-\sin y -1}$$

Now $$\frac{\cos y-\sin y -1}{y}=y\frac{\cos y-1}{y^2}-\frac{\sin y}{y}\rightarrow -1$$ Thus the limit (assuming I have made no error) is $4$.

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Writing $1=\tan\dfrac\pi4,\sqrt2=\sec\dfrac\pi4$

replacing $\dfrac\pi4=2A,x=2y$

$$\lim_{y\to A}\dfrac{\tan^22A-\tan^22y}{\sec2A\cos2y-1}=\cos2A\cdot\lim_{y\to A}(\tan2A+\tan2y)\cdot\lim_{y\to A}\dfrac{\tan2A-\tan2y}{\cos2y-\cos2A}$$

Now $\lim_{y\to A}\dfrac{\tan2A-\tan2y}{\cos2y-\cos2A}$

$=\lim_{y\to A}\dfrac{\sin(2A-2y)}{2\cos2A\cos2y\sin(A+y)\sin(A-y)}$

$=\lim_{y\to A}\dfrac{2\sin(A-y)\cos(A-y)}{2\cos2A\cos2y\sin(A+y)\sin(A-y)}=?$

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Just another way to do it.

Let $x=y+\frac \pi 4$, expand and simplify. You should get $$\lim_{x\to\pi/4} \frac{1-\tan^2(x)}{\sqrt{2}\times \cos(x)-1}=\lim_{y\to 0}\frac{2 (\cos(y)-\sin (y)+1)}{(\cos (y)-\sin (y))^2}$$ which looks very simple.

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