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Limit of a function - Indeterminate Form and LH rule

  • Thread starter Merz
  • Start date
  • #1
2
0

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'Hopitals 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?
 

Answers and Replies

  • #2
354
2
should be -sinx, not +sinx after doing it three times.
But yes, the answer is 2.
 

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