Question

lim x->0 (tanx)^x

Answer 1

\[\large \lim_{x \rightarrow 0} \tan^x x \qquad = \qquad 0^0\]Hmm it appears to give us an indeterminate form.
We'll have to try and manipulate it to get a result.

There is an important trick that we want to take advantage of.
Let me refresh your memory in case you've forgotten.\[\huge \color{royalblue}{e^{\ln x}=x}\]The exponential and the logarithmic function are `inverses` of one another. They sort of... "cancel out" when you do this.




We're going to apply this idea in reverse.
We're going to rewrite our limit in a logarithm, with a base e.\[\huge e^{\ln(\lim_{x \rightarrow 0}\tan^xx)}\]We'll ignore the \(e\) for now and just look at the exponent.
Pulling the limit outside gives us,\[\large \lim_{x \rightarrow 0} \ln(\tan^xx)\]Using rules of logarithms, we can pull the exponent outside as a factor in front of the log.\[\large \lim_{x \rightarrow 0}x \tan x\]

Answer 2

If you haven't learned how to do it this way, then it might be confusing.
It's the same as writing \(y=\) and then taking the log of both sides.
We're doing essentially the same thing.
Maybe you've learned it that way.

Answer 3

We have to employ another weird little algebra trick.
We can rewrite x as it's reciprocal of it's reciprocal.
We flip it, so it's \(\dfrac{1}{x}\) changing our limit to \(\dfrac{\tan x}{x}\).
Then we flip the x again, changing our limit to,\[\lim_{x \rightarrow 0} \frac{\tan x}{\dfrac{1}{x}}\]

Answer 4

This is equivalent to x tan x.

Answer 5

And now maybe you can see that if we take this limit ~
As x approaches 0, our function is approaching \(\dfrac{-\infty}{\infty}\).

This is one of the indeterminate forms that we're allowed to apply `L'Hopital's Rule` to.

Answer 6

Confused? :C yer too quiet lol

Answer 7

Woops, we should have \[\large \lim_{x \rightarrow 0} \frac{\ln(\tan x)}{\dfrac{1}{x}}\]Sorry that was a silly typo, it's giving us -infty/infty though :)

Answer 8

why cant i leave the ln and the e
and rewrite it as e^(xlntanx)
plug in zero
and get e^0
=1

Answer 9

You're getting\[\huge e^{0\cdot -\infty}\]Not e^0

Answer 10

ok thanks