Question

Using L'Hopital's rule, find the limit of...

Answer 1

\[\lim_{x \rightarrow \frac{ \pi ^{+} }{ 2 }}\left( 1 + ctgx \right)^{\frac{ 1 }{ \cos ^{2}x }}\]

Answer 2

So, the limit of (1 + ctgx)^sec²x as x approaches pi/2 from the right, correct?

What is ctgx? I'm not familiar with that function, if that's what it is.

Answer 3

Okay, so ctgx = tanx, right?

Answer 4

ctgx = cotagents = cot = 1/tan

Answer 5

Alright. I have some stuff to do at the moment, but I'll come back to this later. Is that okay?

Answer 6

no problem...

Answer 7

Let \[y = (1 + \cot x)^{\sec^2 x}\]
So, equivalently, you have
\[\lim y\\
\lim e^{\ln y}\\
e^{\lim (\ln y)}\]
(KEEP THIS IN MIND! Whatever the result of the next limit will be, the answer will be e to the power of "limit"!)

So, you must evaluate the following limit:
\[\lim_{x \to \pi/2^+} \ln{(1 + \cot x)}^{\sec^2 x}\\
\lim_{x \to \pi/2^+} \sec^2 x\ln{(1 + \cot x)}\\
\lim_{x \to \pi/2^+} \dfrac{\ln{(1 + \cot x)}}{\cos^2 x}\]
If you need more guidance here, let me know.

Answer 8

where did you get this from:
\[\lim e ^{In y}\]

Answer 9

It's a property of logarithms/exponents.
Say you have y.
y = y, right? That much is clear.
Also, ln(y) = ln(y). Also pretty obvious.
e^(lny) = e^(lny), too.

The property I'm referring to is that, whenever you have a base, like e, raised to the power of a logarithm of the same base, you're left with the number after the logarithm. Simply put,
\[b^{\log_b x} = x\]
(The base of the log is b.)

So, you can see that
e^(lny) = y
The reason for writing the function like this is to drop down the exponent on the log, like I did in my second to last step.

Answer 10

thanks a lot!! i get it.