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Evaluate $$\lim_{x \to \infty}\sqrt x(\sqrt{x+c}- \sqrt x )$$

Attempt:

$$\begin{align} \lim_{x \to \infty}\sqrt x(\sqrt{x+c}- \sqrt x ) &= \lim_{x\to \infty }(\sqrt{x^2+cx}- x) \\ &= \lim _{x\to \infty}x\left(\sqrt{\left(1+\dfrac{c}{x}\right)}-1\right)\\ &= \lim _{x \to \infty} x \times 0 \\ &= 0 \times \infty \\ &=0 \end{align}$$

But the answer given is :

$$\frac c 2$$

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  • $\begingroup$ you can mouseover the pinkish field to reveal $\endgroup$ – Vinyl_cape_jawa Apr 27 '18 at 9:05
  • $\begingroup$ The issue with your reasoning is that you cannot keep one $x$ while making the other term $0$. The two terms go together. The same flawed reasoning would show $1 = \lim_{x \to \infty} x \frac{1}{x} = \lim_{x \to \infty} x 0 = 0$. It makes no sense to throw the limit onto only one term. $\endgroup$ – mathworker21 Apr 27 '18 at 9:12
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Write like this (using $a^2-b^2=(a-b)(a+b)$)

$$ \sqrt{x^2+cx} - x = \frac{ x^2+cx - x^2}{\sqrt{x^2+cx}+x} = \frac{cx}{\sqrt{x^2+cx}+x}$$

Now, factor $x$ numerator and denominator cancel and we obtain

$$ \frac{ c }{ \sqrt{ 1 + \frac{c}{x} } + 1 } $$

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  • $\begingroup$ What have I done wrong? That was the question. $\endgroup$ – Archer Apr 27 '18 at 9:07
  • $\begingroup$ $0*\infty$ is not a valid thing $\endgroup$ – Frank Moses Apr 27 '18 at 9:08
  • $\begingroup$ you have to make simplification to make it a valid thing if it is possible $\endgroup$ – Frank Moses Apr 27 '18 at 9:08
  • $\begingroup$ and I think $0\times \infty$ is not equal to $0$ $\endgroup$ – Frank Moses Apr 27 '18 at 9:09
  • $\begingroup$ why not @FrankMoses $\endgroup$ – Archer Apr 27 '18 at 9:17
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Taylor expansion of $\sqrt{1+f(x)}$ where $f(x)\to 0$ is:

$$\sqrt{1+f(x)}=1+\frac12f(x)+ o(f(x))$$ Then

$$\lim _{x\to \infty}x\left(\sqrt{1+\dfrac{c}{x}}-1\right)=\lim _{x\to \infty}x\left(1+\frac12\dfrac{c}{x}-1\right)=\frac c2$$

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Note $$\displaystyle \lim_{x \to \infty} x = \infty$$ and $$\displaystyle \lim_{x \to \infty} \frac cx = 0$$ while

$$\displaystyle \lim_{x \to \infty} x \cdot\frac cx = c$$

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1) Let $c=0,$ and $x \gt 0$.

The limit is?

2) Let $x > |c|>0.$

$f(x):= x^{1/2}[(x+c)^{1/2}-x^{1/2}] =$

$x^{1/2} \dfrac {c}{(x+c)^{1/2} +x^{1/2}}=$

$\dfrac{c}{(1+c/x)^{1/2} +1}.$

The limit is?

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