Where does it goes?

Calculus Level 2

Find the value of:

$\large \lim\limits_{x \to \frac{\pi}{2}-} (\tan x)^{\cos x}$

Bonus : Prove that $\lim\limits_{x \to \frac{π}{2}+} f(x)$ does not exist.

The answer is 1.

2 solutions

Pi Han Goh
May 13, 2019

Let $\displaystyle y = \lim_{x\to {\frac{\pi}2}^-} (\tan x)^{\cos x}$ . Then $\ln y = \lim_{x\to {\frac{\pi}2}^-} \cos x \cdot \ln(\tan x),$ which is in the indeterminate form of $0 \times \infty$ . L'Hôpital's Rule comes into play! $\ln y = \lim_{x\to {\frac{\pi}2}^-} \frac{ \ln(\tan x) }{ \sec x} = \lim_{x\to {\frac{\pi}2}^-} \frac{ \frac d{dx} \ln(\tan x) }{ \frac d{dx} \sec x} = \lim_{x\to {\frac{\pi}2}^-} \frac{ \frac{\sec^2 x}{\tan x} }{ \sec x \tan x} = \lim_{x\to {\frac{\pi}2}^-} \frac{ \cos x}{\sin^2 x} = 0$ Our answer is $y = e^0 = \boxed 1$ .

For the bonus question:

Let $\displaystyle z = \lim_{x\to {\frac{\pi}2}^+} (\tan x)^{\cos x}$ . Then $\ln z = \lim_{x\to {\frac{\pi}2}^+} \cos x \cdot \ln(\tan x),$ but since $\tan x$ is a negative number for $\frac \pi 2 < x < \pi$ , the natural logarithm of such number does not exist, so the(right hand) limit does not exist.

A Former Brilliant Member
May 9, 2019

The limit is 1 from both directions.

Why is it from both directions? $\displaystyle\lim_{x\rightarrow\frac{\pi}{2}^{+}} (\cos x)^{\cos x}$ is undefined.

Blan Morrison - 2 years, 1 month ago

tan(x)^cos(x)

A Former Brilliant Member - 2 years ago

What do you mean?

Blan Morrison - 2 years ago

The question was about $(\tan{x})^{\cos{x}}$ .

A Former Brilliant Member - 2 years ago

You're mistaken, the left hand limit is defined, but the right hand limit is undefined.

Pi Han Goh - 2 years ago

They are both defined. You have agreed already that from below is defined.

$\lim_{x\to \left(\frac{\pi }{2}\right)^+} \left(\exp \left(\log \left(\tan ^{\cos x} x\right)\right)\right)$

$\lim_{x\to \left(\frac{\pi }{2}\right)^+} \left(\exp \fbox{\$(\cos x) (\log (\tan x))\$}\right)$

$\fbox{\$\exp \left(\lim_{x\to \left(\frac{\pi }{2}\right)^+} ((\cos x) (\log (\tan x)))\right)\$}$

The following step is only to set up an $\frac{\infty}{\infty}$ situation so that L'Hopital rule can be applied.

$\exp \fbox{\$\frac{\lim_{x\to \left(\frac{\pi }{2}\right)^+} (\log (\tan x))}{\frac{1}{\cos x}}\$}$

Added explanation: $\log \left(\tan \left(\frac{\pi }{2}\right)\right) \Rightarrow \infty$ and $\frac{1}{\cos \left(\frac{\pi }{2}\right)}$ is complex infinity.

$\lim_{x\to \left(\frac{\pi }{2}\right)^+}\ \frac{(\log (x \tan ))}{\frac{1}{\cos x}} =$

$\lim_{x\to \left(\frac{\pi }{2}\right)^+}\ \frac{\frac{d}{dx}\log (x \tan )}{\frac{d}{dx}\left(\frac{1}{\cos x}\right)} =$

$\lim_{x\to \left(\frac{\pi }{2}\right)^+}$ $\frac{\frac{\sec ^2}{tan(x)}}{\frac{\sin(x)}{(\cos x)^2}} =$

$\\lim_{x\to \left(\frac{\pi }{2}\right)^+}\ \frac{1}{(\sin x) (\tan x)}$

$\exp \fbox{\$\frac{\lim_{x\to \left(\frac{\pi }{2}\right)^+} \left(\left(\cos ^2 x\right) \left(\sec ^2 x\right)\right)}{(\sin x) (\tan x)}\$}$

$\exp \fbox{\$\frac{\lim_{x\to \left(\frac{\pi }{2}\right)^+} \left(\left(\cos ^2 x\right) \left(\sec ^2 x\right)\right)}{\lim_{x\to \left(\frac{\pi }{2}\right)^+} ((\sin x) (\tan x))}\$}$

$\frac{\exp \fbox{\$\lim_{x\to \left(\frac{\pi }{2}\right)^+} 1\$}}{\lim_{x\to \left(\frac{\pi }{2}\right)^+} ((\sin x) (\tan x))}$

$\frac{\exp 1}{\lim_{x\to \left(\frac{\pi }{2}\right)^+} ((\sin x) (\tan x))}$

$\frac{\exp 1}{\fbox{\$\left(\lim_{x\to \left(\frac{\pi }{2}\right)^+} (\sin x)\right) \left(\lim_{x\to \left(\frac{\pi }{2}\right)^+} (\tan x)\right)\$}}$

$\frac{\exp 1}{\fbox{\$\lim_{x\to \left(\frac{\pi }{2}\right)^+} (\tan x)\$}}$

$e^{\frac{1}{\fbox{\$-\infty \$}}}$

$1$

A Former Brilliant Member - 2 years ago

Your 4th step is wrong. You can't apply L'hopital rule as it is not in an indeterminate form.

Pi Han Goh - 2 years ago

$\frac{\log (\tan x)}{\frac{1}{\cos x}}$ is an indefinite form. It is $\frac{\infty}{\infty}$ . To be precise, the denominator is a complex infinity whose polar form has an infinite magnitude and an unknown phase angle.

$\log \left(\tan \left(\frac{\pi }{2}\right)\right) \Rightarrow \infty$ and $\frac{1}{\cos \left(\frac{\pi }{2}\right)} = \frac10 \Rightarrow \text{ complex }\infty$

A Former Brilliant Member - 2 years ago

@A Former Brilliant Member – If we want to compute the left hand limit or the right hand limit of the function $f(x)$ , then the variable $x$ must only be a real number, so the following limit does not exist (it doesn't even become unboundedly large).

$\lim_{x\to {\frac\pi2}^+ } \ln(\tan x)$

There is no point introducing the complex numbers into the limit.

Pi Han Goh - 2 years ago

@Pi Han Goh – 1) There is nothing in the problem specification to eliminate complex numbers 2) The limit is the same from both directions. This is not a step discontinuity. 3) The fact that one infinity is in fact a complex infinity does not affect the final result.

A Former Brilliant Member - 2 years ago

@A Former Brilliant Member – I think we have a disagreement in convention here. Whenever I see a limit that doesn't have any non-real numbers in it, then I'll only consider the real axis, which there are only two directions to approach a limit. Whereas in the complex plane, there are infinite ways to approach a limit.

Looking at this very question the author posed (with the added bonus question), I'm betting that it came from an introductory calculus text/course, where the functions encountered there are normally real-valued functions, that is, $f: \mathbb R \to \mathbb R$ .

Pi Han Goh - 2 years ago

Where does it goes?

The answer is 1.

2 solutions

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