Welcome to our community

Be a part of something great, join today!

Another Limit, without L'hospital rule

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]
 

chisigma

Well-known member
Feb 13, 2012
1,704
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
Feb 14, 2012
79
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
Feb 13, 2012
1,704

Chipset3600

Member
Feb 14, 2012
79

chisigma

Well-known member
Feb 13, 2012
1,704

Chipset3600

Member
Feb 14, 2012
79
Is $\displaystyle 5^{y}= e^{y\ \ln 5}$...

Kind regards

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