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avatar+191 

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

Any thoughts?

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

 Dec 20, 2018
 #1
avatar+102444 
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+101798 
+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+102444 
+3

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

 

\(\displaystyle\lim_{x\rightarrow\infty}\;\frac{\sqrt{x^4-10x}-\sqrt{x^4-5x^2+7}}{5}\\ =\displaystyle\lim_{x\rightarrow\infty}\;\frac{ \left(\sqrt{x^4-10x}-\sqrt{x^4-5x^2+7}\right) \left(\sqrt{x^4-10x}+\sqrt{x^4-5x^2+7}\right)}{5 \left(\sqrt{x^4-10x}+\sqrt{x^4-5x^2+7}\right)}\\ =\displaystyle\lim_{x\rightarrow\infty}\;\frac{ x^4-10x-(x^4-5x^2+7) }{5 \left(\sqrt{x^4-10x}+\sqrt{x^4-5x^2+7}\right)}\\ =\displaystyle\lim_{x\rightarrow\infty}\;\frac{ 5x^2 -10x-7 }{5 \left(\sqrt{x^4-10x}+\sqrt{x^4-5x^2+7}\right)}\\ =\displaystyle\lim_{x\rightarrow\infty}\;\frac{ 5x^2 -10x-7 }{5 \left(\sqrt{x^4-10x}\right)\left(1+\frac{\sqrt{x^4-5x^2+7}}{\sqrt{x^4-10x}}\right)}\\ \)

 

\( =\displaystyle\frac{1}{5}\times\displaystyle\lim_{x\rightarrow\infty}\;\frac{     5x^2  -10x-7                           }{ \left(\sqrt{x^4-10x}\right)} \times\;\;\displaystyle\lim_{x\rightarrow\infty}\;\frac{  1}{\left(1+\frac{\sqrt{x^4-5x^2+7}}{\sqrt{x^4-10x}}\right)}\\ \)

NOW I will look at each of these limits seperately.

 

\(\displaystyle\lim_{x\rightarrow\infty}\;\frac{     5x^2  -10x-7    }{ \left(\sqrt{x^4-10x}\right)}\\ =\displaystyle\lim_{x\rightarrow\infty}\;\sqrt{\frac{     (5x^2  -10x-7 )^2   }{ x^4-10x}}\\ \text{expanding gives}\\ =\displaystyle\lim_{x\rightarrow\infty}\;\sqrt{\frac{   25x^4-100x^3+30x^2+140x+49   }{ x^4-10x}}\\ \text{Dividing top and bottom by x^4 we get}\\ =\displaystyle\lim_{x\rightarrow\infty}\;\sqrt{\frac{   25-\frac{100}{x}+\frac{30}{x^2}+\frac{140}{x^3}+\frac{49}{x^4}   }{ 1-\frac{10}{x^3}}}\\ =\sqrt{25}\\ =5\\ \text{-------------------------------------------------------}\)

 

\(\displaystyle\lim_{x\rightarrow\infty}\;\frac{  1}{\left(1+\frac{\sqrt{x^4-5x^2+7}}{\sqrt{x^4-10x}}\right)}\\~\\ =\frac{  1}{\displaystyle\lim_{x\rightarrow\infty}\;\left(1+\frac{\sqrt{x^4-5x^2+7}}{\sqrt{x^4-10x}}\right)}\\ =\frac{  1}{\displaystyle\lim_{x\rightarrow\infty}\;\left(\frac{\sqrt{x^4-5x^2+7}}{\sqrt{x^4-10x}}\right)+1}\\ =\frac{  1}{\displaystyle\lim_{x\rightarrow\infty}\;\sqrt{\left(\frac{x^4-5x^2+7}{x^4-10x}\right)}+1}\\ =\frac{  1}{\displaystyle\lim_{x\rightarrow\infty}\;\sqrt{\left(\frac{1-\frac{5}{x^2}+\frac{7}{x^4}}{1-\frac{10}{x^3}}\right)}+1}\\ =\frac{  1}{\displaystyle\lim_{x\rightarrow\infty}\;\sqrt{\left(\frac{1-\frac{5}{x^2}+\frac{7}{x^4}}{1-\frac{10}{x^3}}\right)}+1}\\ =\displaystyle \frac{1}{2}\)

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

 

SO what do we have now.

 

\(\displaystyle\lim_{x\rightarrow\infty}\;\frac{\sqrt{x^4-10x}-\sqrt{x^4-5x^2+7}}{5}= \frac{1}{5} \times5\times \frac{1}{2}=\boxed{\frac{1}{2}}\)

.
 Dec 20, 2018
 #4
avatar+22515 
+8

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

\(\large { \lim \limits_{x \to \infty} \dfrac{\sqrt{x^4-10x}-\sqrt{x^4-5x^2+7}}{5} }\)

 

\(\begin{array}{|rcll|} \hline &&\mathbf{ \lim \limits_{x \to \infty} \dfrac{\sqrt{x^4-10x}-\sqrt{x^4-5x^2+7}}{5} } \\\\ &=&\dfrac15 \cdot \lim \limits_{x \to \infty} \left( \sqrt{x^4-10x}-\sqrt{x^4-5x^2+7} \right) \\\\ &=&\dfrac15 \cdot \lim \limits_{x \to \infty} \left( \sqrt{x^4-10x}-\sqrt{x^4-5x^2+7} \right) \dfrac{\left( \sqrt{x^4-10x}+\sqrt{x^4-5x^2+7} \right)}{\left( \sqrt{x^4-10x}+\sqrt{x^4-5x^2+7} \right)} \\\\ &=&\dfrac15 \cdot \lim \limits_{x \to \infty} \left( \dfrac{ x^4-10x-(x^4-5x^2+7)} { \sqrt{x^4-10x}+\sqrt{x^4-5x^2+7} } \right) \\\\ &=&\dfrac15 \cdot \lim \limits_{x \to \infty} \left( \dfrac{ 5x^2-10x-7} { \sqrt{x^4-10x}+\sqrt{x^4-5x^2+7} } \right) \cdot \dfrac{x^2}{x^2} \\\\ &=&\dfrac15 \cdot \lim \limits_{x \to \infty} \left( \dfrac{ \frac{5x^2-10x-7} {x^2} } { \frac{\sqrt{x^4-10x}+\sqrt{x^4-5x^2+7}} {x^2} } \right) \\\\ &=&\dfrac15 \cdot \lim \limits_{x \to \infty} \left( \dfrac{ \frac{5x^2-10x-7} {x^2} } { \sqrt{\frac{x^4-10x} {x^4} }+\sqrt{ \frac{x^4-5x^2+7} {x^4} } } \right) \\\\ &=&\dfrac15 \cdot \lim \limits_{x \to \infty} \left( \dfrac{ 5-\frac{10}{x}-\frac{7}{x^2} } { \sqrt{1-\frac{10}{x^3} } + \sqrt{1-\frac{5}{x^2}+\frac{7}{x^4}} } \right) \\\\ &=&\dfrac15 \cdot \left( \dfrac{ 5-0-0 } { \sqrt{1-0 } + \sqrt{1-0+0} } \right) \\\\ &=&\dfrac15 \cdot \left( \dfrac{ 5 } { 1 + 1 } \right) \\\\ &=&\dfrac15 \cdot \dfrac{ 5 } {2 } \\\\ &\mathbf{=}&\mathbf{\dfrac12} \\ \hline \end{array}\)

 

 

laugh

 Dec 20, 2018
edited by heureka  Dec 20, 2018
 #5
avatar+101798 
+2

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

 

cool cool cool

 Dec 20, 2018
 #7
avatar+102444 
+1

Thanks Chris.

Melody  Dec 21, 2018
 #6
avatar+191 
+3

Thank you, everyone! 

 Dec 20, 2018

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