Chipset3600 said: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 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
$\chi$ $\sigma$
Chipset3600 said:I can't understood ur z=yln(5)
chisigma said:Simply You set $y\ \ln 5 = z$ and then insert z in (1)...
Kind regards
$\chi$ $\sigma$
Chipset3600 said:where it came from this ln(5)?
chisigma said:Is $\displaystyle 5^{y}= e^{y\ \ln 5}$...
Kind regards
$\chi$ $\sigma$
"Another Limit, without L'hospital rule" is a mathematical concept used to find the limit of a function without using the L'Hospital's rule. This technique involves simplifying the function algebraically and using other limit laws to find the answer.
Learning how to find limits without L'hospital rule allows for a better understanding of mathematical concepts and principles. It also helps in solving more complex problems and in developing critical thinking skills.
The steps to find a limit without using L'hospital rule are as follows:
No, L'hospital rule can only be used in certain types of limit problems, such as indeterminate forms (e.g. 0/0 or ∞/∞). It cannot be used if the limit is not in the form of an indeterminate form.
While L'hospital rule can be a helpful tool in finding limits, it may not always give the correct answer or may be difficult to apply in certain situations. It is important to understand other methods, such as finding limits without L'hospital rule, to verify the answer or to use when L'hospital rule cannot be applied.