MHB Another Limit, without L'hospital rule

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The discussion focuses on finding the limit of the expression (5^x - 25)/(x - 2) as x approaches 2 without using L'Hôpital's rule. The solution involves substituting x - 2 with y, leading to a transformation that incorporates the natural logarithm of 5. By further substituting y with z = y ln(5), the limit simplifies to a fundamental limit involving e^z. Ultimately, the result of the limit is determined to be 25 ln(5). The participants express gratitude for the clarification and successful resolution of the problem.
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Hello MHB, i can't hv success with this limit, help me please:
\lim_{x->2}\frac{5^{x}-25}{x-2}
 
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Chipset3600 said:
Hello MHB, i can't hv success with this limit, help me please:
\lim_{x->2}\frac{5^{x}-25}{x-2}

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

$\chi$ $\sigma$
 
chisigma said:
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 can't understood ur z=yln(5)
 
Chipset3600 said:
I can't understood ur z=yln(5)

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

Kind regards

$\chi$ $\sigma$
 
chisigma said:
Simply You set $y\ \ln 5 = z$ and then insert z in (1)...

Kind regards

$\chi$ $\sigma$

where it came from this ln(5)?
 
Chipset3600 said:
where it came from this ln(5)?

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

Kind regards

$\chi$ $\sigma$
 
chisigma said:
Is $\displaystyle 5^{y}= e^{y\ \ln 5}$...

Kind regards

$\chi$ $\sigma$

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

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