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February 27th, 2019, 08:02 AM  #1 
Newbie Joined: Feb 2019 From: São Paulo, Brasil. Posts: 1 Thanks: 0  Evaluating limits algebraically
Can anyone solve this, please? I'm stuck! \[\lim_{x \rightarrow 0}\frac{\frac{1}{\sqrt{9+x}}\frac{1}{3}}{x}\] 
February 28th, 2019, 06:16 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,663 Thanks: 2642 Math Focus: Mainly analysis and algebra 
We always multiply by the "conjugate" to get rid of roots, using the idea that $$a^2  b^2 = (ab)(a+b)$$ Here, it's perhaps easiest to evaluate the subtraction in the numerator first and simplify: $$\frac1{\sqrt{9+x}}  \frac13 = \frac{3\sqrt{9+x}}{3\sqrt{9+x}}$$ So that $$\frac{\frac1{\sqrt{9+x}}  \frac13}{x} = \frac{\frac{3\sqrt{9+x}}{3\sqrt{9+x}}}{x} = \frac{3\sqrt{9+x}}{3x\sqrt{9+x}}$$ Now, with $a=3$ and $b=\sqrt{9+x}$ we can use the identity above on the numerator. But if we multiply the numerator by $(a+b)=(3+\sqrt{9+x})$, we must also multiply the denominator by that same value to avoid changing the value of the fraction. \begin{align}\frac{3\sqrt{9+x}}{3x\sqrt{9+x}} &= \frac{3\sqrt{9+x}}{3x\sqrt{9+x}} \cdot \frac{3+\sqrt{9+x}}{3+\sqrt{9+x}} \\ &= \frac{9  (9+x)}{{3x\sqrt{9+x}}(3+\sqrt{9+x})} \\ &= \frac{x}{{3x\sqrt{9+x}}(3+\sqrt{9+x})} \\ &= \frac{1}{{3\sqrt{9+x}}(3+\sqrt{9+x})} \end{align} Now you should be able to evaluate the limit directly by setting $x=0$. This works because we cancelled the terms $x$ which were sending both the numerator and denominator to zero. 
February 28th, 2019, 09:01 AM  #3 
Senior Member Joined: Dec 2015 From: somewhere Posts: 534 Thanks: 81 
$\displaystyle x=t^2 9$ , $\displaystyle t\rightarrow 3$ . $\displaystyle \lim_{t\rightarrow 3} \frac{(t3) }{18t(t3)}=1/54 $ . 

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