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How do I evaluate this without using L'Hospital's rule:$$\lim_{x \to 1}\frac {2\cos({x-1})-2}{({x²}-2\sqrt{x}+1 )}\ ?$$

Note : I used L'Hospital's Rule I find $\frac{-4}{3}$ but in wolfram alpha is $0$

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We start by using the double-angle identity for the cosine function to write

$$2\cos (x-1)-2=-4\sin^2\left(\frac{x-1}{2}\right) \tag 1$$

Next, we factor the denominator of the term of interest to find

$$\begin{align} x^2-2x^{1/2}+1&=(x^{1/2}-1)(x^{3/2}+x+x^{1/2}-1)\\\\ &=\frac{(x^{3/2}+x+x^{1/2}-1)(x-1)}{x^{1/2}+1} \tag 2 \end{align}$$

Then, using $(1)$ and $(2)$ we have

$$\frac{2\cos (x-1)-2}{x^2-2x^{1/2}+1}=\left(-4\frac{x^{1/2}+1}{x^{3/2}+x+x^{1/2}-1}\right) \times \left(\frac{\sin\left(\frac{x-1}{2}\right)}{x-1}\right)\times\left( \sin\left(\frac{x-1}{2}\right) \right)\tag 3$$

The first parenthetical term on the right-hand side of $(3)$ approaches $-4$ as $x\to 1$, the second term approaches $1/2$, and the third term approaches $0$.

Finally, we have

$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 1}\left(\frac{2\cos (x-1)-2}{x^2-2x^{1/2}+1}\right)=0}$$

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Set $x-1=u$ and rewrite the expression:

  • $2\cos(x-1)-2=2\cos u-2=-u^2+o(u^2)$,
  • $\begin{aligned}[t]x^2-2\sqrt x+1&=(1+u)^2-2\sqrt{1+u}+1=1+2u+u^2-2-u+\dfrac{u^2}4+o(u^2)+1\\&=u+\dfrac{5u^2}4+o(u^2) \end{aligned}.$

Thus $2\cos(x-1)-2\sim_1 -(x-1)^2$, $\;x^2-2\sqrt x+1\sim_1 x-1$, and $$\dfrac{2\cos(x-1)-2}{x^2-2\sqrt x+1}\sim_1\frac{-(x-1)^2}{x-1}=1-x \xrightarrow[x\to 1]{} 0.$$

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  • $\begingroup$ Solid answer! +1... I believe that asymptotic analysis ought to be taught more often and at an earlier stage in a curriculum. Well done. $\endgroup$ – Mark Viola Oct 2 '15 at 23:04
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Let $f(x) = 2 \cos (x-1), g(x) = x^2 - 2 \sqrt x.$ We are looking at

$$\frac{f(x) - f(1)}{g(x) -g(1)} = \frac{(f(x) - f(1))/(x-1)}{(g(x) -g(1))/(x-1)}.$$

By the definition of the derivative, the above $\to f'(1)/g'(1)$ as $x\to 1.$ This is a simple computation.

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$$\displaystyle \lim_{x \to 1}\frac {2\cos({x-1})-2}{({x²}-2\sqrt{x}+1 )}$$

$$=-2\cdot\dfrac1{\lim_{x\to1}\{1+\cos(x-1)\}}\cdot\left(\lim_{x\to1}\dfrac{\sin(x-1)}{x-1}\right)^2\cdot\lim_{x\to1}\dfrac{(x-1)^2}{x^2-2\sqrt x+1}$$

Set $\sqrt x-1=y\implies x=(y+1)^2$ in $$\lim_{x\to1}\dfrac{(x-1)^2}{x^2-2\sqrt x+1}$$

to get $$\lim_{y\to0}\dfrac{\{(1+y)^2-1\}^2}{(1+y)^4-2(y+1)+1} =\lim_{y\to0}\dfrac{y^2(2+y)^2}{2y+O(y^2)}=\cdots=0$$

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Set $u=\sqrt x$, then $$...=\lim_{u\to 1}\frac{2\cos(u^2-1)-2}{u^4-2u+1}=\lim_{u\to 1}\frac{2\cos(u^2-1)-2}{(u-1)(u^3+u^2+u-1)}.$$

And use the fact that $$\cos(y-1)=1-\frac{(y-1)^2}{2}+o((y-1)^2).$$

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  • $\begingroup$ I used this but i didn't get 0 $\endgroup$ – zeraoulia rafik Oct 2 '15 at 22:08
  • $\begingroup$ Of course it make $0$ since $$2\cos(u^2-1)-2=2-(u^2-1)^2-2+\underbrace{o((u^2-1)^2)}_{=o((u-1)^2)}=-(u-1)^2(u+1)^2+o((u-1)^2).$$ Therefore you can simplify by $u-1$. $\endgroup$ – Surb Oct 2 '15 at 22:17

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