Question

lim as x approaches infinity of xtan(3/x)

Answer 1

God these are giving me so much trouble... okay here's my attempt.

Answer 2

Should I turn it into a fraction before I do lhopitals?
As in, turn tan in to sin/cos???

Answer 3

\[\text{ Let } u=\frac{3}{x} =>\text{ as } x->\infty, u-> 0\]
=> x=3/u
\[\lim_{u \rightarrow 0}\frac{3}{u} \cdot \tan(u)\]
we can apply l'hosptial rule since we have 0/0
\[3 \lim_{u \rightarrow 0}\frac{\sec^2(u)}{1}=3 \sec^2(0)=3(1)=3\]

Answer 4

let \[u=\frac{1}{x} \]
\[\lim_{u \rightarrow 0} \frac{\tan3u}{u} \]using lhopitals rule
\[\lim_{u \rightarrow 0}\frac{3\sec^2u}{1}\]\[=\lim_{u \rightarrow 0}\frac{3}{\cos^2{3u}} =\frac{3}{1} = 3\]

Answer 5

I suppose I don't need to. Lhopitals should work regardless of if it's in fraction form.


Okay so....
xtan(3/x) = infinity * 0 when x is approaching infinity.

That's a prerequisite of lhopitals!

SO i can get right down to business.

So derivative of x is 1. Then derivative of tan(3/x) is sec^2(3/x) *(-3/x^2)

Did i do that correctly ??^^^

Answer 6

@myininaya


I dont get how the hell you got (3/u).... from the (3/u)tan(u)

Answer 7

i let u=3/x
so ux=3
so x=3/u
if u=3/x and x goes to infinity, then u goes to zero