Another Limit, without L'hospital rule

[Chipset3600]

Hello MHB, i can't hv success with this limit, help me please:
[TEX]\lim_{x->2}\frac{5^{x}-25}{x-2}[/TEX]

 

[chisigma]

Hello MHB, i can't hv success with this limit, help me please:
[TEX]\lim_{x->2}\frac{5^{x}-25}{x-2}[/TEX]

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

$\chi$ $\sigma$

 

[Chipset3600]

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

$\chi$ $\sigma$

I cant understood ur z=yln(5)

 

[chisigma]

I cant understood ur z=yln(5)

Simply You set $y\ \ln 5 = z$ and then insert z in (1)...

Kind regards

$\chi$ $\sigma$

 

[Chipset3600]

Simply You set $y\ \ln 5 = z$ and then insert z in (1)...

Kind regards

$\chi$ $\sigma$

where it came from this ln(5)?

 

[chisigma]

where it came from this ln(5)?

Is $\displaystyle 5^{y}= e^{y\ \ln 5}$...

Kind regards

$\chi$ $\sigma$

 

[Chipset3600]

Is $\displaystyle 5^{y}= e^{y\ \ln 5}$...

Kind regards

$\chi$ $\sigma$

Now i understood, nd i find the result 25ln(5). Thank you