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I was just reviewing my calc book when I stumbled across this problem.

Any thoughts?

limxx410xx45x2+75

 Dec 20, 2018
 #1
avatar+118696 
0

Yes, CPhill is right, I entered the wrong function.   Sorry.

 Dec 20, 2018
edited by Guest  Dec 20, 2018
edited by Melody  Dec 20, 2018
 #2
avatar+130466 
+2

The guest keyed in the incorrect function.....the limit = 1/2

 

How to solve this algebraically....I don't know....[maybe a series expansion ???]

 

Here's graphical proof :   https://www.desmos.com/calculator/hqlnsz5lrt

 

 

cool cool cool

CPhill  Dec 20, 2018
 #3
avatar+118696 
+3

\lim_{x\longrightarrow\infty}\frac{\sqrt{x^4-10x}-\sqrt{x^4-5x^2+7}}{5}

 

limxx410xx45x2+75=limx(x410xx45x2+7)(x410x+x45x2+7)5(x410x+x45x2+7)=limxx410x(x45x2+7)5(x410x+x45x2+7)=limx5x210x75(x410x+x45x2+7)=limx5x210x75(x410x)(1+x45x2+7x410x)

 

=15×limx5x210x7(x410x)×limx1(1+x45x2+7x410x)

NOW I will look at each of these limits seperately.

 

limx5x210x7(x410x)=limx(5x210x7)2x410xexpanding gives=limx25x4100x3+30x2+140x+49x410xDividing top and bottom by x^4 we get=limx25100x+30x2+140x3+49x4110x3=25=5-------------------------------------------------------

 

limx1(1+x45x2+7x410x) =1limx(1+x45x2+7x410x)=1limx(x45x2+7x410x)+1=1limx(x45x2+7x410x)+1=1limx(15x2+7x4110x3)+1=1limx(15x2+7x4110x3)+1=12

-------------------------------------------------------------

 

SO what do we have now.

 

limxx410xx45x2+75=15×5×12=12

 Dec 20, 2018
 #4
avatar+26396 
+8

Limits
I was just reviewing my calc book when I stumbled across this problem.

limxx410xx45x2+75

 

limxx410xx45x2+75=15limx(x410xx45x2+7)=15limx(x410xx45x2+7)(x410x+x45x2+7)(x410x+x45x2+7)=15limx(x410x(x45x2+7)x410x+x45x2+7)=15limx(5x210x7x410x+x45x2+7)x2x2=15limx(5x210x7x2x410x+x45x2+7x2)=15limx(5x210x7x2x410xx4+x45x2+7x4)=15limx(510x7x2110x3+15x2+7x4)=15(50010+10+0)=15(51+1)=1552=12

 

 

laugh

 Dec 20, 2018
edited by heureka  Dec 20, 2018
 #5
avatar+130466 
+2

Nice.....Melody and heureka....!!!!!

 

cool cool cool

 Dec 20, 2018
 #7
avatar+118696 
+1

Thanks Chris.

Melody  Dec 21, 2018
 #6
avatar+199 
+3

Thank you, everyone! 

 Dec 20, 2018

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