September 9th, 2017, 06:49 PM  #1 
Member Joined: Apr 2017 From: New York Posts: 41 Thanks: 2  limit
Hi guys, Can you explain me how I can find limit in this question? what method and how? thanks Question: Limit as x is approaching to 0.1 from the left what is the limit of (10x1) / 10x^3x^2 
September 9th, 2017, 07:17 PM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,031 Thanks: 2342 Math Focus: Mainly analysis and algebra 
You should factor the denominator and treat the left and right limits separately. Thus as $x$ approaches $0.1$ from the left you have $10x1 < 0$, while as $x$ approaches $0.1$ from the right you have $10x  1 > 0$.

September 9th, 2017, 07:26 PM  #3 
Member Joined: Apr 2017 From: New York Posts: 41 Thanks: 2 
If I plug in 0.1 then it becomes 0/0 ?

September 9th, 2017, 08:09 PM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,031 Thanks: 2342 Math Focus: Mainly analysis and algebra 
Yes.

September 10th, 2017, 01:29 AM  #5 
Global Moderator Joined: Dec 2006 Posts: 18,142 Thanks: 1417 
As the wording "x is approaching to 0.1 from the left" implies that x < 0.1, (10x  1)/10x³  x² = (10x  1)/x²(10x  1) = 1/x². Now, plug x = 0.1 into 1/x² to get the answer. Is any further explanation of this method needed? 
September 10th, 2017, 05:06 AM  #6 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,820 Thanks: 750 
There are polynomials in both numerator and denominator and the fact that they are both 0 at x= 0.1 means that each has a factor of (x 0.1): 10x 1= 10(x 0.1) and $\displaystyle 10x^3 x^2= 10x^2(x 0.1)$. For x< 0.1 ("from the left"), x 0.1< 0 so $\displaystyle 10x^3 x^2< 0$ and $\displaystyle 10x^3 x^2= 10x^2x 0.1= 10x^2(x 0.1)$. $\displaystyle \frac{10x 1}{10x^3 x^2}= \frac{10(x 0.1)}{10x^2x 0.1}= \frac{10}{10x^2}= \frac{1}{x^2}$ which goes to $\displaystyle \frac{1}{(0.1)^2}= \frac{1}{0.01}= 100$. 

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