Limit problem - calc 1

  • Thread starter hayesk85
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  • #1
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limit problem - calc 1 please see reply

Homework Statement



limx->0 1/x



The Attempt at a Solution



I know this is easy if you graph it on the calculator, but I want to know the logic behind the getting the answer without graphing it.

Thank you
 
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Answers and Replies

  • #2
79
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Well, plug in values less than 1 and more than -1 and you'll see that you get closer and closer to positive or negative infinity. For example 1/.1 = 10, 1/-.1 = -10, 1/.01 = 100, 1/-.01 = -100.

Now if you know enough about limits, then you'll realize that:

[tex]\lim_{x\rightarrow0} \frac{1}{x}[/tex]

does not exist because:

[tex]\lim_{x\rightarrow0^+} \frac{1}{x} [/tex] does not equal [tex]\lim_{x\rightarrow0^-} \frac{1}{x}[/tex]
 
  • #3
7
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Sorry, I meant:

limx->0 sin(1/x)

Now its harder!
 
  • #4
79
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Well technically, you can look at it almost the same way as I described above. There is one other thing to notice though. Remember as x gets closer and closer to zero, then 1/x gets infinitely larger. Now what if 1/x was equal to [itex]\pi/2[/itex] then the sin would be 1. Now what if x got closer to zero and 1/x was equal to [itex]3\pi/2[/itex], then the sin would -1. So, this means as you keep getting closer and closer to zero (from both sides), each side oscillates an infinite amount of times (alternating oscillations on each side of the y axis of course because of negative and positive x's resulting in negative and positive sin's).

I hope that clears your reasoning up a little bit. At least that is how I like to look at it, but that's my first time ever explaining it to someone, so I'm sorry if it sounds a bit confusing.
 
  • #5
45
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Expanding on what chislam has said, you may or may not be aware of the theorom stating that using any sequence of x that tends to 0, (xn), then if the limit exists f((xn)) must converge to this limit for all subsequences. I will leave it to you to find the required subsequences to prove the limit doesn't exist.
 

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