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: 3,092 Thanks: 845 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
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 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

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