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- #1
Chipset3600
Member
- Feb 14, 2012
- 79
Hello MHB, i can't hv success with this limit, help me please:
[TEX]\lim_{x->2}\frac{5^{x}-25}{x-2}[/TEX]
[TEX]\lim_{x->2}\frac{5^{x}-25}{x-2}[/TEX]
Setting $x-2=y$ You obtain...Hello MHB, i can't hv success with this limit, help me please:
[TEX]\lim_{x->2}\frac{5^{x}-25}{x-2}[/TEX]
I cant understood ur z=yln(5)Setting $x-2=y$ You obtain...
$\displaystyle \frac{5^{x}-25}{x-2} = \frac{5^{2+y}-5^{2}}{y}= 5^{2}\ \frac{5^{y}-1}{y} = 5^{2}\ \ln 5\ \frac{e^{y\ \ln 5} -1} {y\ \ln 5}$ (1)
and setting $z= y\ \ln 5$ You arrive at the limit...
$\displaystyle \lim_{z \rightarrow 0} 5^{2}\ \ln 5\ \frac{e^{z}-1}{z}$ (2)
... who contains a 'fundamental limit'...
Kind regards
$\chi$ $\sigma$
Simply You set $y\ \ln 5 = z$ and then insert z in (1)...I cant understood ur z=yln(5)
where it came from this ln(5)?Simply You set $y\ \ln 5 = z$ and then insert z in (1)...
Kind regards
$\chi$ $\sigma$
Is $\displaystyle 5^{y}= e^{y\ \ln 5}$...where it came from this ln(5)?
Now i understood, nd i find the result 25ln(5). Thank youIs $\displaystyle 5^{y}= e^{y\ \ln 5}$...
Kind regards
$\chi$ $\sigma$