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$$\lim_{x\to\pi/4}\frac{ 1-\tan^2x}{\sqrt{2}\cos x -1}$$

I'm trying to find the limit of this function without applying L'Hopital's Rule.

I'm really stuck with this. Any transformation I perform to it ends up giving me a limit of the form $\lim_{x\to\pi/4}f(x) = \frac{0}{0}$.

My last attemt was this:

\begin{align} \frac{1-\tan^2x}{\sqrt{2}\cos x -1} &= \frac{\cos^2 x - \sin^2 x}{\sqrt{2}\cos x -1} \\ &= \frac{(\cos x - \sin x)(\cos x + \sin x)}{\sqrt{2}\cos x -1} \end{align} Then I tried to find a way to get rid of the factors with negative signs but so far I've not been able to do better.

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What about: \begin{eqnarray*} \frac{1 - \tan^2 x}{\sqrt{2}\cos x - 1} &=& \frac{1 - \tan^2 x}{\sqrt{2}\cos x - 1}\frac{\sqrt{2}\cos x + 1}{\sqrt{2}\cos x + 1}\\ &=& (\sqrt{2}\cos x + 1)\frac{2 - \sec^2 x}{2\cos^2 x - 1}\\ &=& \frac{\sqrt{2}\cos x + 1}{\cos^2 x}\frac{2 - \sec^2 x}{2 - \sec^2 x}\\ &=& \frac{\sqrt{2}\cos x + 1}{\cos^2 x} \end{eqnarray*}

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Some ideas:

  • Change the variable $x\mapsto y+\frac{\pi}{4}$.

  • Calculate $$\lim \frac{\sqrt{2}\cos x -1}{x-\frac{\pi}{4}}$$ and $$\lim \frac{1-\tan x}{x-\frac{\pi}{2}}$$

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Notice that $$\frac{1-\tan^2x}{\sqrt{2}\cos x - 1} = \left(1-\frac{\sin^2x}{\cos^2x}\right)\left(\frac{\sqrt{2}\cos x + 1}{2\cos^2x - 1}\right) = \left(\frac{\sqrt{2}\cos x +1}{\cos^2x}\right)\left(\frac{\cos^2x-\sin^2x}{2\cos^2x-1}\right).$$ Since $\cos^2x-\sin^2x = \cos{2x} = 2\cos^2x-1$, it suffices to find $\lim\limits_{x\rightarrow\pi/4}{\frac{\sqrt{2}\cos x+1}{\cos^2x}}$, which should not be hard.

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Write $f(x) = \tan^2 x, g(x) = \sqrt 2\cos x.$ Then our expression equals

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

As $x\to \pi/4,$ the above $\to -f'(\pi/4)/g'(\pi/4)$ by the definition of the derivative. This is simple enough to compute (and we are not using L'Hopital).

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