Question

lim x->0 (1+tan(x)^1/x

Answer 1

I assume that is supposed to be
\[\lim_{x \rightarrow 0} (1+\tan(x))^{1/x}\]

Often a good thing to explore is changing it into some sort of fraction and using L'Hopital's rule. You could take the natural log and raise e to the resulting log which would give you \[\huge{{\lim_{x\rightarrow 0}(e^{\frac{\ln (1+\tan(x)) }{x}})}}\] and looks promising.

Answer 2

i'm really bad with logarithms, etc..
but i'm guessing you multiplied by e and ln because they cancel each other out, then by multiplying with ln, the exponent can be "taken down" and multiplied by the function?

Answer 3

No, I was taking advantage of the fact that

\[e^{\ln x} = x\]

Answer 4

\[\begin{align*}\left(1+\tan x\right)^{1/x}&=\exp\bigg[\ln(1+\tan x)^{1/x}\bigg]\\
&=\exp\bigg[\frac{1}{x}\ln(1+\tan x)\bigg]\\
&=\exp\bigg[\frac{\ln(1+\tan x)}{x}\bigg]
\end{align*}\]
as per @whpalmer4's suggestion.

If you're not familiar with the notation,
\[\large e^{f(x)}=\exp[f(x)]\]
Then, using the continuity of the exponential function at 0, you can change the limiting function:
\[\lim_{x\to0}\exp\bigg[\frac{\ln(1+\tan x)}{x}\bigg]=\exp\bigg[\lim_{x\to0}\frac{\ln(1+\tan x)}{x}\bigg]\]

Answer 5

Yeah, that's much easier to read with the exp notation, thanks!

Answer 6

after you apply L'Hopital, the fraction ought to reduce to something you can directly evaluate at \(x=0\) and then you've just got \(e\) to that power as your limit.

Answer 7

Oh, also \[\ln u^n = n \ln u\]makes an appearance to generate that fraction: \[u = 1+\tan x\]and\[n =1/x\]

Answer 8

thanks!
But how come i can change the limit function?

Answer 9

and after i get limx→0exp[ln(1+tanx)x] i can apply lhopital?

Answer 10

I don't understand the "exp" notation... What is exp?

Answer 11

@Jhannybean

\[\exp(x) = e^{x}\]

Answer 12

it's just different notation

Answer 13

Oh okay.

Answer 14

Thanks to my web browser for discarding a nearly complete post without confirmation :-(

We had as the guts of our limit

\[(1+\tan x)^{1/x}\] which we can rewrite without changing its value as\[(1/x) \ln (1+\tan x)\] or\[\frac{\ln(1+\tan x)}{x}\]

The reason we can do that is thanks to two properties of logarithms:

\[\exp[{\ln u}] = u\]\[\ln u^n = n \ln u\]

Finally, the limit of a constant (e) to a power is equal to the constant to the limit of the power, so our limit becomes

\[\exp[\lim_{x->0} \frac{1+\tan x}{x} ]\]and we can use L'Hopital on the fraction.

Answer 15

\[\frac{d}{dx}[1+\tan x] = \sec^2 x\]\[\frac{d}{dx}[x] = 1\]

so our limit after L'Hoping is

\[\exp[\lim_{x->0} \sec^2x]\]which can be evaluated directly.

Answer 16

alright, thank you all for your help

Answer 17

What do you get for your answer?

Answer 18

hmm, where did ln go?

Answer 19

exp[limx−>0 1+tanx/x], after exp[ln 1+tanx/x] ?

Answer 20

because of exp[lnu]=u this propery?

Answer 21

property*

Answer 22

but then the exp would also go away..

Answer 23

Oh, sorry, I oversimplified there, didn't I :-)

\[\frac{d}{dx}[\ln (1+\tan x)] = \frac{\sec^2x}{1+\tan x}\]by the chain rule.

Answer 24

oh right, forgot about chain rule

Answer 25

i'm still confused, which step is that?

Answer 26

doing L'Hopital....

Answer 27

that's the derivative of the numerator, and the derivative of the denominator is just 1

Answer 28

in any case, sec^2 x/(1+tan x) can be evaluated at x = 0

Answer 29

i understand everything to: exp {ln(1+tanx)/x}
Using: exp {ln u} = u
I should get: (1+tanx)/x , right?

Answer 30

oh wait, u= 1+tanx, i get it

Answer 31

but i would think my 'u' changed to 1+tanx/x ....

Answer 32

remember, the limit is inside the exp[ ]...

Answer 33

\(\large \cfrac{\ln(1+\tan(x))}{x}\) using LH rule : \(\large \cfrac{f'(x)}{g'(x)}\)\[\large \cfrac{\cfrac{1\cdot \sec^2(x)}{1+\tan(x)}}{1} = \cfrac{\sec^2(x)}{1+\tan(x)}\]

Answer 34

so we take the derivative of ln (1+tan x) over the derivative of x as we do L'Hopital's rule, and that gives us sec^2 x/(1+tan x) which we evaluate at x = 0. tan (0) = 0, sec (0) = 1, so we have exp[1^2/(1+0)] = exp[1] = ...

Answer 35

And @whpalmer4 just explained word per word my evaluation of the function. :)

Answer 36

sp the answer should just be e

Answer 37

so*

Answer 38

yep, eeeeeeeee the magical number.....

Answer 39

and i use the quotient rule to derive

Answer 40

quotient rule? chain rule, isn't it? you're taking the derivative of ln u where u = 1 + tan x

Answer 41

i see

Answer 42

\[\huge e^{\lim_{x \rightarrow 0}\cfrac{\sec^2(x)}{1+\tan(x)}} = e^{\cfrac{\sec^2(0)}{1+\tan(0)}} = e^{\cfrac{1}{1}}=e\]

Answer 43

lol. hideous.....but readable.

Answer 44

nice, thanks