Solving a Limit: Finding the Right Approach

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SUMMARY

The limit problem discussed is \lim_{x \rightarrow \infty} \left( 1 + 10^{-x}\right)^{10^x}. This limit can be solved by recognizing its similarity to the well-known limit \lim_{x \rightarrow \infty} \left( 1 + \frac{1}{x} \right) ^ x = e. By applying this concept, the limit can be evaluated effectively, leading to the conclusion that the original limit approaches the value of e as x approaches infinity.

PREREQUISITES
  • Understanding of limits in calculus
  • Familiarity with exponential functions
  • Knowledge of the mathematical constant e
  • Basic logarithmic properties and their applications
NEXT STEPS
  • Study the derivation of the limit \lim_{x \rightarrow \infty} \left( 1 + \frac{1}{x} \right) ^ x = e
  • Explore advanced limit techniques, including L'Hôpital's Rule
  • Investigate the applications of the number e in calculus and real analysis
  • Learn about sequences and series convergence related to limits
USEFUL FOR

Students of calculus, mathematics educators, and anyone interested in advanced limit evaluation techniques.

twoflower
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Hi all,

could you help me how to solve this limit?

[tex] \lim_{x \rightarrow \infty} \left( 1 + 10^{-x}\right)^\left(10^x\right)[/tex]

Common adjustments for log involving the limit log (1)/0 don't work and no other idea comes to my mind...

Thank you for any help.
 
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Hmm,...
Look at the problem again, it looks incredibly like:
[tex]\lim_{x \rightarrow \infty} \left( 1 + \frac{1}{x} \right) ^ x = e[/tex]
Can you go from here?
Viet Dao,
 
VietDao29 said:
Hmm,...
Look at the problem again, it looks incredibly like:
[tex]\lim_{x \rightarrow \infty} \left( 1 + \frac{1}{x} \right) ^ x = e[/tex]
Can you go from here?
Viet Dao,

Yep, you're right, can't believe I didn't see it :) Thanks.
 

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