Another Limit, without L'hospital rule

  • Context: MHB 
  • Thread starter Thread starter Chipset3600
  • Start date Start date
  • Tags Tags
    Limit
Click For Summary
SUMMARY

The discussion focuses on evaluating the limit \(\lim_{x \to 2} \frac{5^{x}-25}{x-2}\) using substitution techniques. The user successfully transforms the limit by setting \(x-2=y\) and subsequently \(z=y \ln 5\), leading to the expression \(\lim_{z \to 0} 5^{2} \ln 5 \frac{e^{z}-1}{z}\). This approach reveals the fundamental limit, resulting in the final answer of \(25 \ln(5)\). The discussion emphasizes the importance of understanding exponential functions and their logarithmic properties in limit evaluations.

PREREQUISITES
  • Understanding of limits in calculus
  • Familiarity with exponential functions and logarithms
  • Knowledge of substitution techniques in limit problems
  • Basic comprehension of the fundamental limit \(\lim_{z \to 0} \frac{e^{z}-1}{z} = 1\)
NEXT STEPS
  • Study the properties of exponential functions and their derivatives
  • Learn about the fundamental limit and its applications in calculus
  • Explore advanced limit techniques, including L'Hôpital's Rule and Taylor series expansions
  • Practice solving limits involving logarithmic and exponential functions
USEFUL FOR

Students and educators in calculus, mathematicians focusing on limit evaluations, and anyone interested in mastering techniques for solving complex limits involving exponential functions.

Chipset3600
Messages
79
Reaction score
0
Hello MHB, i can't hv success with this limit, help me please:
\lim_{x->2}\frac{5^{x}-25}{x-2}
 
Physics news on Phys.org
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 :)
 

Similar threads

MHB Calculating the limits without the L'Hospital rule
  • · Replies 2 ·
Replies
2
Views
2K
MHB How Can I Solve This Limit Without L'Hospital's Rule?
  • · Replies 3 ·
Replies
3
Views
2K
MHB How Can You Solve This Limit Problem Without L'Hospital's Rule?
  • · Replies 2 ·
Replies
2
Views
2K
MHB Derivative using the limit definition (without using L'Hospital's rule)
  • · Replies 4 ·
Replies
4
Views
2K
High School Rationale Behind t-Substitution for Evaluating Limits?
  • · Replies 5 ·
Replies
5
Views
2K
MHB Finding Limit using L'Hospital's Rule - Need Help!
  • · Replies 2 ·
Replies
2
Views
2K
High School Proof of Chain Rule: Understanding the Limits
  • · Replies 3 ·
Replies
3
Views
2K
MHB Limits with the natural logarithm
  • · Replies 3 ·
Replies
3
Views
2K
Undergrad A limit as t-->0 of log(t) / SQRT(t)
  • · Replies 17 ·
Replies
17
Views
2K
MHB Limit involving a hyperbolic function
  • · Replies 4 ·
Replies
4
Views
2K