Hii! please help asap. i’ll give brainliest
1 answer:
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Answer:
-1
Step-by-step explanation:
The expression evaluates to the indeterminate form -∞/∞, so L'Hopital's rule is appropriately applied. We assume this is the common log.
d(log(x))/dx = 1/(x·ln(10))
d(log(cot(x)))/dx = 1/(cot(x)·ln(10)·(-csc²(x)) = -1/(sin(x)·cos(x)·ln(10))
Then the ratio of these derivatives is ...
lim = -sin(x)cos(x)·ln(10)/(x·ln(10)) = -sin(x)cos(x)/x
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At x=0, this has the indeterminate form 0/0, so L'Hopital's rule can be applied again.
d(-sin(x)cos(x))/dx = -cos(2x)
dx/dx = 1
so the limit is ...
lim = -cos(2x)/1
lim = -1 when evaluated at x=0.
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I find it useful to use a graphing calculator to give an estimate of the limit of an indeterminate form.
