Calculating Limit: How to Show Lim x^2-sin(x)→∞

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SUMMARY

The limit of the function x^2 - sin(x) as x approaches infinity is confirmed to be infinity. By applying the Squeeze Theorem, it is established that since sin(x) oscillates between -1 and 1, the function is bounded by x^2 - 1 and x^2 + 1. As x tends to infinity, both bounding functions approach infinity, thereby confirming that lim x^2 - sin(x) also approaches infinity.

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  • Understanding of limits in calculus
  • Familiarity with the Squeeze Theorem
  • Basic knowledge of trigonometric functions, specifically sin(x)
  • Concept of asymptotic behavior of functions
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How do I formally show that lim x^2 - sin(x) as x tends to infinity is infinity?
 
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I suppose you could look at the minimum possible value of your function... since sin(x) can never be greater than one, your equation can never be less than x^2-1. So work with that instead.
 
There's a theorem that is sometimes called the Squeeze Theorem or Squeeze Play Theorem. If f(x) <= g(x) <= h(x) for all x in a suitable domain, and lim f(x) = lim h(x) = L, then lim g(x) = L.

x^2 - 1 <= x^2 - sin(x) <= x^2 + 1 for all x. What can you say about lim x^2 -1 and lim x^2 + 1 as x grows large without bound?
 

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