My Math Forum Hi...I'm a new member here...I would be glad if anyone could see the following is cor

 Calculus Calculus Math Forum

 March 2nd, 2017, 01:54 AM #1 Newbie   Joined: Mar 2017 From: US Posts: 2 Thanks: 0 Hi...I'm a new member here...I would be glad if anyone could see the following is cor Shenaya Perera
 March 2nd, 2017, 04:17 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,199 Thanks: 873 The left side of that is rather blurry and hard to read but I believe you have $\lim_{x\to\infty} \frac{(x+ 1)^{2012}+ (x+ 2)^{2012}+ \cdot\cdot\cdot+ (x+ 2017)^{2012}}{x^{2012}+ 2017^{2012}}$ You then divide both numerator and denominator by $x^{2012}$, dividing each term in both numerator and denominator by that. Yes, since the numerator is a finite sum, you can do that. Each term in the numerator is $\frac{(x+ a)^{2012}}{x^{2012}}= \left(\frac{x+ a}{x}\right)^{2012}= \left(1+ \frac{a}{x^{2012}\right)^{2012}$ getting $\lim_{x\to\infty} \frac{(1+ 1/x)^{2012}+ (1+ 2/x)^{2012}+ \cdot\cdot\cdot+ (1+ 2017/x)^{2012}}{1+ \left(\frac{2017}{x}\right)^{2012}}$. Now, again because the numerator is a finite sum with 2017 terms, no matter what x is, you can take the limit in each term as x goes to infinity. Of course, each fraction, with x in the denominator, goes to 0 leaving the sum of 2017 "1"s in the numerator while the denominator goes to 1. The result is 2017. Thanks from greg1313 and pererashenaya7
March 2nd, 2017, 04:24 AM   #3
Newbie

Joined: Mar 2017
From: US

Posts: 2
Thanks: 0

Quote:
 Originally Posted by Country Boy The left side of that is rather blurry and hard to read but I believe you have $\lim_{x\to\infty} \frac{(x+ 1)^{2012}+ (x+ 2)^{2012}+ \cdot\cdot\cdot+ (x+ 2017)^{2012}}{x^{2012}+ 2017^{2012}}$ You then divide both numerator and denominator by $x^{2012}$, dividing each term in both numerator and denominator by that. Yes, since the numerator is a finite sum, you can do that. Each term in the numerator is $\frac{(x+ a)^{2012}}{x^{2012}}= \left(\frac{x+ a}{x}\right)^{2012}= \left(1+ \frac{a}{x^{2012}\right)^{2012}$ getting $\lim_{x\to\infty} \frac{(1+ 1/x)^{2012}+ (1+ 2/x)^{2012}+ \cdot\cdot\cdot+ (1+ 2017/x)^{2012}}{1+ \left(\frac{2017}{x}\right)^{2012}}$. Now, again because the numerator is a finite sum with 2017 terms, no matter what x is, you can take the limit in each term as x goes to infinity. Of course, each fraction, with x in the denominator, goes to 0 leaving the sum of 2017 "1"s in the numerator while the denominator goes to 1. The result is 2017.
Sorry for the inconveniece caused and thanks for the answer and your explanation,now I'm sure I've done this in the correct way!

Shenaya Perera

 Tags cor, glad, herei, hiim, member

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post babaliaris New Users 1 November 21st, 2015 03:39 PM Crease124 New Users 2 February 27th, 2013 10:39 AM No-Op New Users 5 October 13th, 2012 08:54 AM The Chaz New Users 20 May 11th, 2011 09:45 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top