Question

Please answer this>>> i get e^1 what my mistake.>>>

Answer 1

\[ \lim_{x \rightarrow 0} sinx^{tanx}\]

Answer 2

whats the answer?

Answer 3

the answer is 1.. it mean e^0

Answer 4

i got 1 as well

Answer 5

how it is. the question ask you to use L'Hopital..

Answer 6

well hopefully somebody can explain how to do it by that method. I am not sure

Answer 7

You get an indeterminate form if you plug 0 in.
0^0 I think.

Hmm let's try using L'Hop. To do so we'll have to revisit an old trick.\[\huge \lim_{x \rightarrow 0} \sin x^{\tan x} \qquad = \qquad \lim_{x \rightarrow 0}e^{\ln(\sin x^{\tan x})}\]

Answer 8

Understand what I did there? I exponentiated AND took the log of the problem, which didn't change it's value.

Answer 9

We want to ignore the e for right now, we'll just look at the inside part.
Using a familiar rule of logarithms we can write it as,\[\huge \lim_{x \rightarrow 0} \color{#CC0033}{\tan x} \color{#3366CF}{\ln(\sin x)}\]

Answer 10

Confused about anything yet? D:

Answer 11

ok.. then. for the left side ln(y)=tanxln(sinx)

Answer 12

Sure if you want to do it that way, that works also :)

Answer 13

so can i chnge tanx =SINX/cosx

Answer 14

let's not. All we've done so far is rearrange things a bit.. we should STILL be getting an indeterminate form.

As we approach 0 now, we'll see that our limit is aproaching \(0( -\infty)\)

Answer 15

We need the indeterminate form, \(\huge \frac{0}{0}\) or \(\huge \frac{\infty}{\infty}\) in order to apply L'Hop Rule.

Answer 16

So from here we can ummm take advantage of a little trick.
Let's rewrite our Tangent function as \(\huge \frac{1}{cot x}\)

Answer 17

\[\huge \lim_{x \rightarrow 0}\frac{\color{#3366CF}{\ln(\sin x)}}{\color{#CC0033}{\cot x}}\]

Answer 18

Does that make sense how we changed it to cotangent? :o

Answer 19

f'(x) = cosx/sinx
g,(x)=-cosec^2(x)

Answer 20

Yah looks good so far c:

Answer 21

\[\large \lim_{x \rightarrow 0} \frac{\frac{\cos x}{\sin x}}{-\csc^2 x} \quad = \quad \lim_{x \rightarrow 0}\frac{-\cos x \sin^2 x}{\sin x}\]
Looks like we can simplify it down a little bit.

Answer 22

can x/0 because sin 0 = 0

Answer 23

You need to simplify before plugging 0 in :O

Answer 24

Cancel stuff out.

Answer 25

ok.. so no more sin xbecause we cancel each other..

Answer 26

what do you understand about the convergent and divergent???

Answer 27

\[\large \lim_{x \rightarrow 0} -\cos x \cdot \sin x\]And I think this should give us a nice answer which is NOT of indeterminate form.
If you're unsure, you could also apply the sine double angle to simplify it down alittle further.

Answer 28

Likes with power series? I'm not very good with those to be honest :C

Answer 29

as know convergent is lim> infinite is =0 if not zero it is divergent>>