# Another Limit, without L'hospital rule

#### Chipset3600

##### Member
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

##### Well-known member
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

##### Member
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

##### Well-known member
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

##### Member
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

##### Well-known member
where it came from this ln(5)?
Is $\displaystyle 5^{y}= e^{y\ \ln 5}$...

Kind regards

$\chi$ $\sigma$

#### Chipset3600

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

Kind regards

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