lim x->0 (sinx)^x
lim (sinx)^x , as x -->0
= 1
would it be possible to get the steps? or some sort of explanation?
I used an intuitive approach.
as x --> 0 , sinx ---> 0 ,but not quite
anything raised to the zero is 1
(except 0^0 is undefined)
try this on your calculator ....
set it to radians
take sin(.000001)^.000001 to get .999986184
take sin(.00000001)^.00000001 to get .999999815
the closer we get to zero, the closer the result gets to 1
thank you!
or, take log:
ln(limit) = x * ln(sin x)
If t = 1/x, we have
ln(sin 1/t) / t
Now take derivatives a couple of times and you wind up with
ln(limit) = 0, so
limit = 1
To find the limit of the expression lim x->0 (sinx)^x, you can use the concept of exponential functions and logarithms.
Step 1: First, rewrite the expression using logarithms. Take the natural logarithm (ln) of both sides of the equation:
ln[lim x->0 (sinx)^x] = ln[lim x->0 sinx^x]
Step 2: Move the exponent x in front of the logarithm using the logarithmic property:
ln[lim x->0 (sinx)^x] = x * ln[lim x->0 sinx]
Step 3: Now consider the limit of the inner expression lim x->0 sinx. Since sinx approaches 0 as x approaches 0, we have:
ln[lim x->0 (sinx)^x] = x * ln[0]
Step 4: Notice that ln[0] is undefined. However, we can use L'Hopital's rule to evaluate the limit. Apply L'Hopital's rule by taking the derivative of the numerator and the denominator with respect to x:
ln[lim x->0 (sinx)^x] = limit x->0 (ln[0]) / (1/x)
Now, we have an indeterminate form of "0/∞". To evaluate this limit further, we can rewrite it as the reciprocal of the limit:
ln[lim x->0 (sinx)^x] = limit x->0 (1/x) / (1/ln[0])
Since ln[0] is undefined and approaching negative infinity, the reciprocal will approach 0:
ln[lim x->0 (sinx)^x] = limit x->0 (1/x) / 0
Step 5: We can rewrite the limit as the reciprocal of the limit:
ln[lim x->0 (sinx)^x] = limit x->0 0 / (1/x)
The reciprocal of zero is infinity, and since (1/x) approaches infinity as x approaches 0, we have:
ln[lim x->0 (sinx)^x] = limit x->0 0 / ∞
Finally, when the numerator is zero and the denominator is infinity, the limit will be zero:
ln[lim x->0 (sinx)^x] = 0
Therefore, the limit of the expression lim x->0 (sinx)^x is 0.