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$$\lim\limits_{x\to 0}\dfrac{\sqrt[4]{x^4+1}-\sqrt{x^2+1}}{x^2}$$

What is the limit? I've tried turning the $4$th root function into a nested square root function in hopes of having an $a^2 - b^2$ in the problem but I'm still clueless on what I'll do next.

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This is the closest thing I have to what you were thinking of.

Notice that:

$$\left(\sqrt[4]{x^4+1}-\sqrt{x^2+1} \right) \left(\sqrt[4]{x^4+1}+\sqrt{x^2+1} \right) \left(\sqrt{x^4+1}+x^2+1 \right)$$ $$= \left(\sqrt{x^4+1} - (x^2+1)\right) \left(\sqrt{x^4+1}+x^2+1 \right)$$ $$= x^4+1 - (x^4+2x^2+1) = -2x^2$$

Therefore the limit becomes:

$$\lim\limits_{x\to 0} \dfrac{\sqrt[4]{x^4+1}-\sqrt{x^2+1}}{x^2} \cdot\dfrac{\sqrt[4]{x^4+1}+\sqrt{x^2+1}}{\sqrt{x^4+1}+\sqrt{x^2+1}} \cdot \dfrac{\sqrt{x^4+1}+x^2+1}{\sqrt{x^4+1}+x^2+1} $$

$$= \lim\limits_{x\to 0} \frac{-2x^2}{x^2 \left( \sqrt{x^4+1}+\sqrt{x^2+1} \right) \left(\sqrt{x^4+1}+x^2+1 \right)} $$

$$= \lim\limits_{x\to 0} \frac{-2}{ \left( \sqrt{x^4+1}+\sqrt{x^2+1} \right) \left(\sqrt{x^4+1}+x^2+1 \right)} $$

and substituting $0$ in, we get that the limit is equal to:

$$\frac{-2}{(1+1)(1+1)} = -\frac{1}{2}$$

The other answers give an easier solution.

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Using binomial series, $$\lim\limits_{x\to 0}\dfrac{\sqrt[4]{x^4+1}-\sqrt{x^2+1}}{x^2}$$

$$=\lim\limits_{x\to0}\dfrac{\left(1+\dfrac14x^4+O(x^8)\right)-\left(1+\dfrac12x^2+O(x^4)\right)}{x^2}$$

$$=\lim\limits_{x\to0}\dfrac{-\dfrac12x^2+O(x^4)}{x^2}=-\dfrac12$$

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    $\begingroup$ +1 Just what I was about to answer. $\endgroup$ – Toby Mak Sep 1 at 5:15
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Let $a=\sqrt[4]{x^4+1}, b=\sqrt{x^2+1}$ $$a-b = {a^2-b^2 \over a+b} = {a^4-b^4 \over (a+b)(a^2+b^2)}$$ $\begin{align} \displaystyle{\lim_{x \to 0}}{\sqrt[4]{x^4+1}-\sqrt{x^2+1} \over x^2} &= \displaystyle{\lim_{x \to 0}}{(x^4+1)-(x^4+2x^2+1) \over x^2(a+b)(a^2+b^2)} \cr &= \displaystyle{\lim_{x \to 0}}{-2 \over (a+b)(a^2+b^2)} \cr &= {-2 \over (1+1)(1+1)} \cr &= -{1\over2} \end{align}$

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  • $\begingroup$ +1 Nice to see your idea on solving the question as well. $\endgroup$ – Toby Mak Sep 1 at 6:05
  • $\begingroup$ Can I ask what kind of approach you're going with this one? $\endgroup$ – Gabriel Kenneth Mariñas Sep 1 at 10:17
  • $\begingroup$ I keep multiply the conjugates to open up the radicals. a^4-b^4 has open-up enough as a simple polynomial -2x^2. Denominator conjugates are all nonzero, a+b = a^2+b^2 = 2. This avoided 0/0 problem $\endgroup$ – albert chan Sep 1 at 10:47
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Interpreted the question as 4th root of $(x^4 + 1)$ and square root of $(x^2+1)$ So, It is of the form 0/0 so I think it will be valid for applying the L'Hôpital's rule. Differentiate it 2 times - we get the ans as $(-1/2)$.

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