Another Limit, without L'hospital rule

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Discussion Overview

The discussion revolves around evaluating the limit \(\lim_{x \to 2} \frac{5^{x}-25}{x-2}\) without using L'Hôpital's rule. Participants explore different approaches and transformations to simplify the limit expression.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Homework-related

Main Points Raised

  • One participant presents a transformation by setting \(x-2=y\) to rewrite the limit, leading to the expression \(\frac{5^{2+y}-5^{2}}{y}\).
  • This transformation allows for further manipulation, resulting in the expression \(5^{2} \ln 5 \frac{e^{y \ln 5} - 1}{y \ln 5}\).
  • A later reply questions the substitution \(z = y \ln 5\) and seeks clarification on its origin.
  • Another participant explains that \(5^{y} = e^{y \ln 5}\), which clarifies the use of \(\ln 5\) in the limit transformation.
  • Ultimately, one participant claims to have understood the process and arrives at the result \(25 \ln(5)\), thanking others for their help.

Areas of Agreement / Disagreement

While there is a progression in understanding among participants, the discussion does not reach a consensus on the method's validity or the correctness of the final result, as some participants express confusion at various points.

Contextual Notes

There are unresolved aspects regarding the transformations used, particularly the justification for introducing \(\ln 5\) and the implications of the fundamental limit referenced.

Chipset3600
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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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