l'Hospital's rules problem

[ijd5000]

Hi,

i need to find the limit as x->infinity of x tan(9/x). It's a type 0*infinity so i rewrote as [tan(9/x)]/1/x and as x/cot(9/x). The first way produced a limit that appears will always be 0/0 and the second way produced (infinity/9). The answer to the problem is '9'.

 

[HallsofIvy]

Writing the function as "tan(9/x)/(1/x)" and using L'Hopital's rule, as you say you did, what did you get for the derivatives of tan(9/x) and (1/x)? When I do that, I do NOT get "0/0".

 

[DrPhil]

Hi,

i need to find the limit as x->infinity of x tan(9/x). It's a type 0*infinity so i rewrote as [tan(9/x)]/1/x and as x/cot(9/x). The first way produced a limit that appears will always be 0/0 and the second way produced (infinity/9). The answer to the problem is '9'.

Another way to rewrite the expression is to let \(\displaystyle u = 1/x\). Then the limit becomes

\(\displaystyle \displaystyle \lim_{u \to 0}\left[ \dfrac{\tan(9u)}{u} \right] \)

That looks a lot "cleaner" to me.

 

[ijd5000]

Writing the function as "tan(9/x)/(1/x)" and using L'Hopital's rule, as you say you did, what did you get for the derivatives of tan(9/x) and (1/x)? When I do that, I do NOT get "0/0".

-(9 sec^2(9/x))/x^2

 

[ijd5000]

Another way to rewrite the expression is to let \(\displaystyle u = 1/x\). Then the limit becomes

\(\displaystyle \displaystyle \lim_{u \to 0}\left[ \dfrac{\tan(9u)}{u} \right] \)

That looks a lot "cleaner" to me.

i get 9 from that without plugging (1/x) back in do I need to?

 

[DrPhil]

i get 9 from that without plugging (1/x) back in do I need to?

Either "x" or "u" is a dummy variable that isn't needed any more after we take the limit. Since there is no "u" in the final result, there is nothing to back-substitute.