Limit of a function - Indeterminate Form and LH rule

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SUMMARY

The discussion centers on evaluating limits involving indeterminate forms using L'Hôpital's Rule. The first limit, as x approaches 0 for the expression (Sin(x) + Cos(x) + e^-x - 2) / (Sin(x) - x), is confirmed to be 2 after applying L'Hôpital's Rule three times. The second limit, as x approaches infinity for (sqrt(x^2 + 1) - 1)^ln(x), is identified as an indeterminate form of infinity minus infinity, requiring further manipulation for resolution. The final conclusion is that the limit evaluates to 2.

PREREQUISITES
  • Understanding of limits in calculus
  • Familiarity with L'Hôpital's Rule
  • Knowledge of indeterminate forms
  • Basic properties of trigonometric and exponential functions
NEXT STEPS
  • Study L'Hôpital's Rule applications in depth
  • Learn about different types of indeterminate forms
  • Explore limit evaluation techniques for logarithmic functions
  • Investigate the behavior of functions as they approach infinity
USEFUL FOR

Students studying calculus, particularly those focusing on limits and L'Hôpital's Rule, as well as educators seeking to clarify these concepts for their students.

Merz
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Homework Statement




Lim x-> 0 Sin(x) + Cos(x) + e^-x - 2 / Sin(x) - x



The Attempt at a Solution



Sin(x) + Cos(x) + e^-x - 2 = 0 and Sin(x) - x = 0

This is indeterminate form?

I used l'hospital's rule 3 times to get

-Cos(x) + Sin(x) - e^-x / -Cos(x)

The top equals -2 and the bottom equals -1. The limit is 2. Am I understanding this right?

Homework Statement



Lim x->infinity (sqrt(x^2+1) - 1)^ln(x)

The Attempt at a Solution



I'm not sure what to do with ln(x). I believe this is the indeterminate form infinity - infinity.
Am I supposed to rewrite this someway?
 
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should be -sinx, not +sinx after doing it three times.
But yes, the answer is 2.
 

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