How can I calculate the limit without using the L'Hopital's rule

Question

$$ \lim_{x \to 2} \frac{1-\sqrt{1-\sqrt{4x-8}}}{1-\sqrt{1-\sqrt{\frac{x-2}{x+2}}}} $$

I have tried to multiply with conjugates or use auxiliary variables but it did not arrive at all simple

Answer 1

HINT:

We need $x\to2^+$

$$\lim_{x \to 2}\frac{1-\sqrt{1-\sqrt{4x-8}}}{1-\sqrt{1-\sqrt{\dfrac{x-2}{x+2}}}}=\lim_{x \to 2}\dfrac{2\sqrt{x-2}}{\sqrt{\dfrac{x-2}{x+2}}}\cdot\dfrac{1+\sqrt{1-\sqrt{\dfrac{x-2}{x+2}}}}{1+\sqrt{1-\sqrt{4x-8}}} $$

Cancel out $\sqrt{x-2}$ as $\sqrt{x-2}\ne0\iff x\ne2$ as $x\to2$

Can you take it from here?

Answer 2

HINT: multiply numerator and denominator by $$1+\sqrt{1-\sqrt{\frac{x-2}{x+2}}}$$ and by $$1+\sqrt{1-\sqrt{4x-8}}$$ and simplify the term

Answer 3

First use $t=x-2$ substitution, to send it to a limit at point $0$: $$ \lim_{x \to 2} \frac{1-\sqrt{1-\sqrt{4x-8}}}{1-\sqrt{1-\sqrt{\frac{x-2}{x+2}}}} \ =\ \lim_{t\to 0^+}\frac{1-\sqrt{1-2\sqrt{t}}}{1-\sqrt{1-\sqrt{\frac{t}{t+4}}}}=\dots$$ Now multiply the numerator and denominator by $1+\sqrt{1-\sqrt{\frac{t}{t+4}}}$ and we get $$\dots=\lim_{t\to 0^+}\frac{\left(1-\sqrt{1-2\sqrt{t}}\,\right)\left(1+\sqrt{1-\sqrt{\frac{t}{t+4}}}\right)}{\sqrt{\frac{t}{t+4}}}$$ The next step: observe that the crucial part is $$\lim_{t\to 0^+}\frac{\left(1-\sqrt{1-2\sqrt{t}}\,\right)}{\sqrt{t}}$$ because the rest tends to $4$ as $t\to 0$ (since in the rest you can just put $t=0$ to obtain the limit).

It is a much easier limit now, can you finish from here?

Answer 4

First, let $x=u+2$ to get

$$\lim_{u\to0}\frac{1-\sqrt{1-\sqrt{4u}}}{1-\sqrt{1-\sqrt{\frac u{u+4}}}}$$

The numerator may be handled with binomial expansion:

$$\sqrt{1-2\sqrt u}=1-\sqrt u+\mathcal O(u)$$

And likewise the denominator:

$$\sqrt{1-\sqrt{\frac u{u+4}}}=1-\frac12\sqrt{\frac u{u+4}}+\mathcal O\left(\frac u{u+4}\right)$$

which leads us to conclude that

$$\lim_{u\to0}\frac{1-\sqrt{1-\sqrt{4u}}}{1-\sqrt{1-\sqrt{\frac u{u+4}}}}=\lim_{u\to0}\frac{\sqrt u+\mathcal O(u)}{\frac12\sqrt{\frac u{u+4}}+\mathcal O\left(\frac u{u+4}\right)}=\lim_{u\to0}\frac{1+\mathcal O(\sqrt u)}{\frac1{2\sqrt{u+4}}+\mathcal O\left(\frac{\sqrt u}{u+4}\right)}=\frac1{\frac1{2\sqrt4}}=4$$

Answer 5

$$\lim_{x \to 2} \frac{1-\sqrt{1-\sqrt{4x-8}}}{1-\sqrt{1-\sqrt{\frac{x-2}{x+2}}}}$$

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Let's let $z = x-2$ (or $x = z + 2$). Then we get

$$\lim_{z \to 0} \frac{1-\sqrt{1-2\sqrt z}}{1-\sqrt{1-\sqrt{\frac{z}{z+4}}}}$$

\begin{align} 1-\sqrt{1-2\sqrt z} &= \dfrac{1-(1-2\sqrt z)}{1+\sqrt{1-2\sqrt z}} \\ &= \dfrac{2\sqrt z}{1+\sqrt{1-2\sqrt z}} \end{align}

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\begin{align} \frac{1}{1-\sqrt{1-\sqrt{\frac{z}{z+4}}}} &=\frac{1+\sqrt{1-\sqrt{\frac{z}{z+4}}}} {1-\left(1-\sqrt{\frac{z}{z+4}}\right)} \\ &=\frac{1+\sqrt{1-\sqrt{\frac{z}{z+4}}}} {\sqrt{\frac{z}{z+4}}} \\ &=\frac{\sqrt{z+4}+\sqrt{z+4-\sqrt{z(z+4)}}} {\sqrt{z}} \\ \end{align}

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\begin{align} \dfrac{1-\sqrt{1-\sqrt{4x-8}}} {1-\sqrt{1-\sqrt{\frac{x-2}{x+2}}}} &= \dfrac{2\sqrt z} {1+\sqrt{1-2\sqrt z}}\cdot \dfrac{\sqrt{z+4}+\sqrt{z+4-\sqrt{z(z+4)}}} {\sqrt{z}}\\ &= 2\dfrac{\sqrt{z+4}+\sqrt{z+4-\sqrt{z(z+4)}}} {1+\sqrt{1-2\sqrt z}} \end{align}

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If we now let $z=0$, we get

\begin{align} \lim_{x \to 2} \frac{1-\sqrt{1-\sqrt{4x-8}}} {1-\sqrt{1-\sqrt{\frac{x-2}{x+2}}}} &= 2 \lim_{z \to 0} \dfrac{\sqrt{z+4}+\sqrt{z+4-\sqrt{z(z+4)}}} {1+\sqrt{1-2\sqrt z}} \\ &= 2 \lim_{z \to 0} \dfrac{2+2}{1+1} \\ &= 4 \\ \end{align}

Answer 6

Let $u = (4x-8)^{1/2}, v = [(x-2)/(x+2)]^{1/2}.$ The expression equals

$$\frac{u}{v}\left (\frac{1-(1-u)^{1/2}}{u}\right )\big /\left(\frac{1-(1-v)^{1/2}}{v}\right ).$$

Now $u/v = 2(x+2)^{1/2} \to 4$ as $x\to 2^+.$ Because both $u,v \to 0$ as $x\to 2^+,$ both fractions in parentheses approach the same nonzero limit, as you can verify by using the conjugate trick, or simply by recognizing the definition of the derivative of $(1-u)^{1/2}$ at $0.$ It follows that the limit is $4\cdot 1=4.$