Going to the Hospital

Calculus Level 4

Let $f(x) = x e^2$ and $g(x) = x^{\ln x},$ and let $\alpha$ and $\beta$ with $\alpha < \beta$ be the two roots of $f(x)-g(x) = 0.$ Also, let

$l= \lim_{x \rightarrow \beta} \ \dfrac{f(x)-c\beta}{g(x)-\beta^2}.$

Then what is the value of $c-l?$

Note: You may use the approximation $e \approx 2.7183 .$

Try my set .

The answer is 7.139.

2 solutions

Ayush Verma
Mar 26, 2015

$f\left( x \right) =g\left( x \right) \Rightarrow { e }^{ 2 }x={ x }^{ \ln { x } }\\ \\ for\quad g\left( x \right) \quad to\quad be\quad defined\quad x>0\Rightarrow { x }^{ \ln { x } }>0\\ \\ so\quad we\quad can\quad take\quad ln\quad of\quad both\quad side\\ \\ \Rightarrow 2+\ln { x={ \left( \ln { x } \right) }^{ 2 } } \Rightarrow \ln { x=-1,2 } \\ \\ \Rightarrow x={ e }^{ -1 },{ e }^{ 2 }\Rightarrow \alpha ={ e }^{ -1 },\beta ={ e }^{ 2 }\quad (as\quad \beta >\alpha )\\ \\ \Rightarrow l={ lim }_{ x\rightarrow { e }^{ 2 } }\cfrac { f\left( x \right) -c{ e }^{ 2 } }{ g\left( x \right) -{ e }^{ 4 } } \\ \\ As\quad f\left( { e }^{ 2 } \right) =g\left( { e }^{ 2 } \right) ={ e }^{ 4 }\\ \\ as\quad denominator\rightarrow 0\quad so\quad numerator\quad should\quad \\ \\ also\quad tends\quad to\quad zero\quad to\quad limit\quad to\quad be\quad existed.So\\ \\ \Rightarrow c{ e }^{ 2 }={ e }^{ 4 }\Rightarrow c={ e }^{ 2 }\\ \\ Now\quad lim\quad is\quad of\quad \left( \cfrac { 0 }{ 0 } \right) form\\ \\ \Rightarrow l={ lim }_{ x\rightarrow { e }^{ 2 } }\cfrac { f^{ ' }\left( x \right) }{ g^{ ' }\left( x \right) } ={ lim }_{ x\rightarrow { e }^{ 2 } }\cfrac { { e }^{ 2 } }{ g\left( x \right) .\cfrac { 2\ln { x } }{ x } } \\ \\ =\cfrac { { e }^{ 2 } }{ { e }^{ 4 }.\cfrac { 2\ln { { e }^{ 2 } } }{ { e }^{ 2 } } } =0.25\\ \\ \Rightarrow c-l={ e }^{ 2 }-0.25=7.139\\$

Thanks. I have updated the answer to 7.139.

Calvin Lin Staff - 6 years, 2 months ago

I am really sorry for the inconvinience.

Parth Lohomi - 6 years, 2 months ago

This answer makes one mistake, I think: In order to solve for $c$ , it assumes the limit $l$ exists (we shouldn't assume that). A way to explicitly solve for $c$ is...

$l = \lim_{x\rightarrow e^2} \frac{f(x) - ce^2}{g(x) - e^4} \implies$

$l \lim_{x\rightarrow e^2} (g(x) - e^4) = \lim_{x\rightarrow e^2} f(x) - ce^2$

You know the limit on the left-side is zero, which makes it easy to solve for $c$

J P - 2 years, 1 month ago

If we do not assume that limit exists then the Left hand side would not exist which means Right hand side would not exist.

Meet gera - 1 month, 1 week ago

Tanishq Varshney
Mar 26, 2015

the answer is wrong Parth Lohomi . it should be $c-\frac{1}{l}$ then only u get 3.389. do correct me if i am wrong..

exactly the correct answer is comming up to be 7.1390.

Soumya Dubey - 6 years, 2 months ago

Thanks. I have updated the answer to 7.139.

In future, if you spot any errors with a problem, you can “report” it by selecting "report problem" in the “dot dot dot” menu in the lower right corner. This will notify the problem creator who can fix the issues.

Calvin Lin Staff - 6 years, 2 months ago

cool .thanks :)

Soumya Dubey - 6 years, 2 months ago

Going to the Hospital

The answer is 7.139.

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