Please help will give brainllest!
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.
In the given question, the two sides of the given triangles are parallel, so the the two corresponding angles of the Triangles are equal to one another.
Therefore, the Triangles are similar by AA similarity criteria ~
And since they are similar, the ratio of sides of one Triangle will be equal to ratio of corresponding sides of the other Triangle.
that is ~
Hence, value of l = 15.4 units ~