Matematické Fórum

stuart clark · 19. 02. 2015 18:04

Evaluation of Limit $\lim_{x\rightarrow 2} \left(-5\right)^x\;,$ please explain me in detail

Answer given as $\bf{\Limit\; does\; not\; exists}$

Thanks

Pavel · 19. 02. 2015 19:10

↑ stuart clark:

The real exponential function $a^x$ is defined only for $a>0$ and $a\neq 1$ . It means that expressions such as

$(-5)^{1.8},\ (-5)^{2.001}$

etc. are not defined. The function $(-5)^x$ is not defined in any neighborhood of the point $x_0=2$ .

Therefore the given limit does not exist.

Freedy · 19. 02. 2015 21:49

Look at definition of the limit. You are not able to find $\delta >0$ , because the function is not defined in neighborhood of this point as ↑ Pavel: said.

Bati · 19. 02. 2015 22:19

But what if we translated it as a complex function: $(-5)^x=5^x(\cos{\pi x}+i\sin{\pi x})$ ? Then the limit would make perfect sense.

stuart clark · 23. 02. 2015 14:28

Thanks ↑ Pavel:↑ Bati:.

To ↑ Bati: would you like to explain me further.

Freedy · 08. 06. 2015 02:16

Hi,

then we would get $\lim_{x\to2}5^x(\cos \pi x+\text{i}\sin \pi x)$ now we have the real part and imaginary one
(R) $\lim_{x\to2}5^x\cos \pi x=25$
(I) $\lim_{x\to2}5\text{i}\sin \pi x=0$
So I think, that ↑ Bati: thought this