Limit of e to the tangent of x

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SUMMARY

The limit of \( e^{\tan(x)} \) as \( x \) approaches \( \frac{\pi}{2}^{+} \) is definitively 0. As \( x \) approaches \( \frac{\pi}{2} \) from the right, \( \tan(x) \) approaches \( -\infty \). Consequently, \( e^{\tan(x)} \) approaches \( e^{-\infty} \), which equals 0. This conclusion is supported by the behavior of the tangent function and its graph near the vertical asymptote at \( x = \frac{\pi}{2} \).

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thereidisanceman
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lim {e}^{tan(x)}
x\implies\{(pi/2)}^{+}

How do I start? Do I plug in and if so what next?
 
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I would first consider what we get for:

$$\lim_{x\to\frac{\pi}{2}^{+}}\tan(x)$$

What is the value of the above limit?
 
tan (\pi/2)
 
thereidisanceman said:
tan (\pi/2)

Think of the graph of:

$$y=\tan(x)$$

Where is the curve going as we approach $$x=\frac{\pi}{2}$$ from the right side?
 
oh, -\infty from the right
 
thereidisanceman said:
oh, -\infty from the right

Yes, and so what is $e$ raised to this power?

Note: to get your $\LaTeX$ code to parse correctly, wrap it in the MATH tags. Click the $$\sum$$ button on our toolbar, and then place your code in between the generated tags. :D
 
0?

$$\sin\left({\lim_{{}\to{a}}}\right)$$
(not related)
 
Yes, because:

$$\lim_{x\to-\infty}e^x=0$$ :D
 
Oh coolio, thanks!
 
  • #10
Also, if you right-click on any correctly displaying $\LaTeX$ math, choose Show Math As -> TeX Commands, you will see what someone typed into get a particular output. You can also use single-dollar signs to enclose displayed math, like this: $\int_{-\infty}^{\infty}e^{-x^2} \, dx=\sqrt{\pi}$, which I typed up using

Code:
$\int_{-\infty}^{\infty}e^{-x^2} \, dx=\sqrt{\pi}$,

or you can use double-dollar signs to get a displayed equation like this:

$$\int_{-\infty}^{\infty}e^{-x^2} \, dx=\sqrt{\pi},$$ which I typed up using

Code:
$$\int_{-\infty}^{\infty}e^{-x^2} \, dx=\sqrt{\pi}.$$

Notice the difference in size of the various symbols.
 
  • #11
$$\lim_{x\to\frac{\pi}{2}^+}e^{\tan(x)}$$Over the interval $\left(\frac{\pi}{2},\pi\right]$ we have $e^{\tan(x)}=\frac{1}{e^{|\tan(x)|}}$, hence$$\lim_{x\to\frac{\pi}{2}^+}e^{\tan(x)}=\lim_{x\to\frac{\pi}{2}^+}\frac{1}{e^{|\tan(x)|}}=0$$
 
  • #12
Also , you can see the graph of the function where (tanx = - infinity) when it approaches n(pi/2) or -n(pi\2), therefore the function will go to zero.
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