Question

lim x tends to 0+ (sin x)^x=??

Answer 1

\[\lim_{x \rightarrow 0+} ( sinx)^{x}=??\]

Answer 2

.......

Answer 3

1

Answer 4

Well, \[\lim_{x \rightarrow 0+} x^x = 1\]
and we know that sin x / x --> 1 as x --> 0, so intuitively we would expect that
\[\lim_{x \rightarrow 0+} (\sin x)^x = 1\]

Now to prove it.

Answer 5

??how?

Answer 6

so

Answer 7

First take the ln of the expression and we have
\[\ln( (\sin x)^x ) = x \ln(\sin x) = { \ln(\sin x) \over 1/x}\]
Now numerator and denominator go to zero as x --> 0+. Let's apply l'Hopital's rule; then this last term is equal to
\[{ \cos x / \sin x \over -1/x^2} = x \cos x . {x \over \sin x}\]

Answer 8

Now the limit of each of these three terms respectively is 0, 1 and 1 hence
\[\lim_{x \rightarrow 0+} \ln((\sin x)^x) = 0\]

Answer 9

sorry, dropped the minus sign, but the result is the same.

Answer 10

Now. If
\[\lim_{x \rightarrow 0+} f(x) = A\]
then
\[lim_{x \rightarrow 0+} e^{f(x)} = e^A

Answer 11

\[lim_{x \rightarrow 0+} e^{f(x)} = e^A \]

Answer 12

and so you're done.

Answer 13

srry i am not allowed to use l hospital rule

Answer 14

In that case, bound sin x above and below by some functions, f(x) and g(x) such that
\[\lim_{x \rightarrow 0+} f(x)^x = \lim_{x \rightarrow 0+} g(x)^x = 1.\]
For example, for 0 < x < 1, \[\sin x \leq x \]

Answer 15

and we know that x^x has limit 1 as x -->0+

Answer 16

Now you need a lower bound for sin x, g(x) where you can calculate the limit of g(x)^x

Answer 17

oh cmplicated

Answer 18

For example g(x) = x/10 will work. I.e., for 0 < x < 1
\[{x \over 10} \leq \sin x \leq x\]
hence
\[\left({x \over 10}\right)^x \leq (\sin x)^x \leq x^x\]

Answer 19

and limit of both functions on either side as x --> 0+ is 1. Hence the limit we want is 1.

Answer 20

what happened to your picture btw?

Answer 21

?

Answer 22

your profile pic

Answer 23

it was good. Anyway, this is how I think you have to calculate your limit.

Answer 24

isnt the pic good?

Answer 25

the new one? It's fine.

Answer 26

But the old one was better if you ask me. But anyway.

Answer 27

:)

Answer 28

hey James can you help me see where I messed up? The question if from sonofa_nh, you can scroll down to find it.

Answer 29

\[\lim_{x \rightarrow 0} e ^{(1/x*\log x)}\]

Answer 30

=??

Answer 31

post as new question

Answer 32

k

Answer 33

@James: I am not sure, but should we necessarily need to apply L'hospital here?

\[ \ln((\sin x)x)=x \ln(\sin x) \]

now, applying the limits to zero we directly could write the whole thing as zero,since anything multiplied to zero gives zero Isn't?

Answer 34

but ln(0) is undefined, so no, right

Answer 35

No, because ln(sin x) --> -infinity and we don't know for sure that the x is going to cancel it out fast enough

Answer 36

Assuming you know that:
\[ \lim_{x\to0} \frac{\sin{x}}{x} = 1 \quad \text{and} \quad \lim_{x\to0^{+}} x \ln{x} = 0\] You can solve it easier. First note that: \[(\sin{x})^{x} = \exp{(\ln{(\sin{x})^x})} = \exp{(x \ln{(\sin{x})})} \] Then multiply the exponent with (sin x / sin x) resulting in:
\[ \exp{(\frac{x}{\sin{x}} \sin{x} \ln{(\sin{x})})} \to \exp{(1\times0)} = e^{0} =1 \quad \text{when} \quad x \to 0^{+} \]