anonymous
  • anonymous
What is the limit as x approaches infinity of the function sin(x+1/x) and how do you solve that?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\lim_{x\to\infty}\sin\left(x+\frac{1}{x}\right)~~~\text{or}~~~\lim_{x\to\infty}\sin\frac{x+1}{x}~~~?\]
anonymous
  • anonymous
\[\lim_{x \rightarrow \infty}\sin \left( x +\frac{ 1 }{ x } \right)\]
anonymous
  • anonymous
As \(x\to\infty\), do you understand why \(x\to\infty\) and \(\dfrac{1}{x}\to0\) ?

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anonymous
  • anonymous
yes. I think what is sin, because as 1/x approaches infinity is staying at 1 and the denominator is blowing up. x is going on to infinity...right?
anonymous
  • anonymous
I'm not sure I understand what you're saying... My point is that as \(x\to\infty\), you essentially have \(\sin\left(x+\dfrac{1}{x}\right)=\sin x\). So you have to find \(\displaystyle\lim_{x\to\infty}\sin x\).
anonymous
  • anonymous
\[\lim_{x \rightarrow x} \sin x =1\] right?
anonymous
  • anonymous
No, that's not it. As x gets larger and larger, \(\sin x\) will constantly alternate between -1 and 1, without resting at a specific value. What does that tell you about the limit?
anonymous
  • anonymous
the limit does not exist because it's not approaching a specific value
anonymous
  • anonymous
Right!
anonymous
  • anonymous
Thank you!

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