Limits and integration problem

[Krushnaraj Pandya]

Homework Statement

if f(x)= lim(n→∞) e^(xtan(1/n)log(1/n)) and ∫f(x)/(sin^11x.cosx)^1/3 dx=g(x)+c, then
1) g(pi/4)=3/2
2) g(x) is continuous for all x
3) g(pi/4)= -15/8
4) g(pi/4)=12

2. The attempt at a solution
Part a-Evaluating the limit, since 1/n tends to 0, log(1/n)→-∞=-n, using tan(1/n)/(1/n)=1, I get f(x)=e^(-x)
Part b-I tried substituting -x=t, which gives -∫e^t/(-sin^11t.cost)...I'm stuck here, would appreciate some insight, also let me know if part a is correct (I'm not very confident)

 

[Ray Vickson]

Homework Statement

if f(x)= lim(n→∞) e^(xtan(1/n)log(1/n)) and ∫f(x)/(sin^11x.cosx)^1/3 dx=g(x)+c, then
1) g(pi/4)=3/2
2) g(x) is continuous for all x
3) g(pi/4)= -15/8
4) g(pi/4)=12

2. The attempt at a solution
Part a-Evaluating the limit, since 1/n tends to 0, log(1/n)→-∞=-n, using tan(1/n)/(1/n)=1, I get f(x)=e^(-x)
Part b-I tried substituting -x=t, which gives -∫e^t/(-sin^11t.cost)...I'm stuck here, would appreciate some insight, also let me know if part a is correct (I'm not very confident)

No, no, no! You absolutely cannot say that $\log (1/n) \to -\infty = -n$. There are different "levels" of infinity. For example, $n/(n^2+n) \to 0$, even though both the numerator and denominator $\to \infty.$

In fact, $(1/n) \log(1/n) = (1/n) (- \log n) = - \log(n)/n.$ You can evaluate that limit using l'Hospital's rule.

 

[Krushnaraj Pandya]

No, no, no! You absolutely cannot say that $\log (1/n) \to -\infty = -n$. There are different "levels" of infinity. For example, $n/(n^2+n) \to 0$, even though both the numerator and denominator $\to \infty.$

In fact, $(1/n) \log(1/n) = (1/n) (- \log n) = - \log(n)/n.$ You can evaluate that limit using l'Hospital's rule.

ohh, right...I really forgot to revise the chapters I did (sigh!). so -log(n)/n comes out to be -(1/n) which is 0 so f(x) tends to 1, correct?

 

[Krushnaraj Pandya]

ohh, right...I really forgot to revise the chapters I did (sigh!). so -log(n)/n comes out to be -(1/n) which is 0 so f(x) tends to 1, correct?

following that logic, In the integration part I wrote 1/(sin^(11/3)*cos^(1/3), then multiplied and divided by sin^3(x)cos^3(x) which gave cot^2(x)cosec^8(x), putting cosecx=t, it reduced to integral of t^7 - t^9, yet it doesn't give the correct answer
(PS-sorry I didn't write using LaTex, I need some time to learn it)

 

[Ray Vickson]

ohh, right...I really forgot to revise the chapters I did (sigh!). so -log(n)/n comes out to be -(1/n) which is 0 so f(x) tends to 1, correct?

Correct.