My Math Forum  

Go Back   My Math Forum > High School Math Forum > Pre-Calculus

Pre-Calculus Pre-Calculus Math Forum

LinkBack Thread Tools Display Modes
February 27th, 2019, 08:02 AM   #1
Joined: Feb 2019
From: São Paulo, Brasil.

Posts: 1
Thanks: 0

Lightbulb Evaluating limits algebraically

Can anyone solve this, please? I'm stuck!
\[\lim_{x \rightarrow 0}\frac{\frac{1}{\sqrt{9+x}}-\frac{1}{3}}{x}\]
thecousinleo is offline  
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 $$a^2 - b^2 = (a-b)(a+b)$$

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})}

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.
v8archie is offline  
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{-(t-3) }{18t(t-3)}=-1/54 $ .
idontknow is offline  

  My Math Forum > High School Math Forum > Pre-Calculus

algebraically, evaluating, limits

Thread Tools
Display Modes

Similar Threads
Thread Thread Starter Forum Replies Last Post
Need help!! On evaluating limits (matching) orangerify Pre-Calculus 3 September 21st, 2016 03:46 AM
Evaluating Limits nealed1 Calculus 6 March 13th, 2016 07:01 PM
Question about finding limits algebraically. kekie Calculus 1 February 18th, 2014 11:45 PM
Evaluating limits Calc12 Calculus 7 April 1st, 2011 09:04 AM
evaluating limits momo67 Calculus 1 August 31st, 2008 05:22 AM

Copyright © 2019 My Math Forum. All rights reserved.