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 September 9th, 2017, 05:49 PM #1 Member   Joined: Apr 2017 From: New York Posts: 63 Thanks: 6 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 (10x-1) / |10x^3-x^2|
 September 9th, 2017, 06:17 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,235 Thanks: 2412 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 $10x-1 < 0$, while as $x$ approaches $0.1$ from the right you have $10x - 1 > 0$.
 September 9th, 2017, 06:26 PM #3 Member   Joined: Apr 2017 From: New York Posts: 63 Thanks: 6 If I plug in 0.1 then it becomes 0/0 ?
 September 9th, 2017, 07:09 PM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,235 Thanks: 2412 Math Focus: Mainly analysis and algebra Yes.
 September 10th, 2017, 12:29 AM #5 Global Moderator   Joined: Dec 2006 Posts: 18,707 Thanks: 1530 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, 04:06 AM #6 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,967 Thanks: 807 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 $\displaystyle |10x^3- x^2|= 10x^2|x- 0.1|= -10x^2(x- 0.1)$. $\displaystyle \frac{10x- 1}{|10x^3- x^2|}= \frac{10(x- 0.1)}{10x^2|x- 0.1|}= -\frac{10}{10x^2}= -\frac{1}{x^2}$ which goes to $\displaystyle -\frac{1}{(0.1)^2}= -\frac{1}{0.01}= -100$.

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