Limits (1 Viewer)

[coyazayo]

Hi every1 thnx in advance

1.Evaluate the limit

lim x->∞ (xtan(2/x))


2. Use L'Hopital rule to evaluate the following

lim x->∞ (-x^2/e^x)

 

[seanieg89]

First one: 2
tan(t)/t -> 1 as t -> 0. This is a direct application of that.

Second one: 0
e^x grows much faster than any polynomial as x gets large, so the denominator "wins". To prove this using L'Hopital, just differentiate numerator and denominator twice so your numerator becomes a constant. Of course this method will work with the -x^2 replaced by an arbitrary polynomial, just differentiate till it becomes a constant.

 

[eyeseeyou]

Is L'hospital rule in the HSC?

 

[Green Yoda]

Is L'hospital rule in the HSC?

nope

 

[braintic]

Is L'hospital rule in the HSC?

Not only is it not in the course, if you use it without justification you will be penalised.

 

[KingOfActing]

Not only is it not in the course, if you use it without justification you will be penalised.

Quick L'Hopital cheat for 0/0 limits:

 

[seanieg89]

Not that the above is not quite proving the full 0/0 case of L'Hopital, it is proving the weaker assertion that if f,g are diffble on an interval containing c and g'(c) is nonzero then f(x)/g(x)-> f'(c)/g'(c).

The usual formulation of L'Hopital's is that if f,g are diffble on a interval except possibly at an interior point c, and if f'(x)/g'(x)-> L as x->c, then f(x)/g(x) must also tend to L.

This is a bit more general as can be seen by functions involving things like square roots, which are not differentiable at 0, but you could still look at the limiting behaviour of the derivative as you approach 0.

The weaker formulation would probably suffice for pretty much any limit you would need to compute in MX2 though, just something to keep in mind.