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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: 7,691 Thanks: 2670 Math Focus: Mainly analysis and algebra We always multiply by the "conjugate" to get rid of roots, using the idea that $$a^2 - b^2 = (a-b)(a+b)$$ 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: 742 Thanks: 98 $\displaystyle x=t^2 -9$ , $\displaystyle t\rightarrow 3$ . $\displaystyle \lim_{t\rightarrow 3} \frac{-(t-3) }{18t(t-3)}=-1/54$ . Tags algebraically, evaluating, limits Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post orangerify Pre-Calculus 3 September 21st, 2016 03:46 AM nealed1 Calculus 6 March 13th, 2016 07:01 PM kekie Calculus 1 February 18th, 2014 11:45 PM Calc12 Calculus 7 April 1st, 2011 09:04 AM momo67 Calculus 1 August 31st, 2008 05:22 AM

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