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Two statements about limits

  • Thread starter vbplaya
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  • #1
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Hey, I just need help with these two statements about limits which happens to be my least favorite topic of calculus. I just need to know if they are true or false and why.

If lim x→c =L, then f(c) = L.

I don't think that is true because f(c) may not always equal L? or is the statement true? I'm not sure


Also:
If f(x) < g(x) for all x≠a, then lim x→a f(x) < lim x→a g(x)

I'm not sure about this one either. I know that if f(x)=g(x) for all x≠c in an open interval containing c, and their limits exists, then lim x→c f(x) = lim x→c g(x). But I don't know if it's is the same for this inequality.
 

Answers and Replies

  • #2
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I think they're both correct. I'm no master with limits either. They are an awkward concept to grasp, but you should take time and fight the repulsion, because limits are the foundation of calculus. You'll be able to solve run of the mill problems without them, but in order to really understand what's going on, you've got to try to understand them.

That said, I didn't really understand them until the second half of calc 2 heh.
 
  • #3
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vbplaya said:
Hey, I just need help with these two statements about limits which happens to be my least favorite topic of calculus. I just need to know if they are true or false and why.

If lim x→c =L, then f(c) = L.

I don't think that is true because f(c) may not always equal L? or is the statement true? I'm not sure


Also:
If f(x) < g(x) for all x≠a, then lim x→a f(x) < lim x→a g(x)

I'm not sure about this one either. I know that if f(x)=g(x) for all x≠c in an open interval containing c, and their limits exists, then lim x→c f(x) = lim x→c g(x). But I don't know if it's is the same for this inequality.
Both of your statements are false.

*Statement #1 counterexample:
[tex] \text{Let} \; f ( x ) = \frac{{x^2 + x - 6}}{{x - 2}} [/tex]
Then,
[tex] \mathop {\lim }\limits_{x \to 2} f\left( x \right) = \mathop {\lim }\limits_{x \to 2} \frac{{x^2 + x - 6}}{{x - 2}} = \mathop {\lim }\limits_{x \to 2} x + 3 = 5 [/tex]

but the removable discontinuity
[tex] f\left( 2 \right){\text{ does not exist}}{\text{.}} [/tex]

*Statement #2 counterexample:
[tex] {\text{Let }}f\left( x \right) = 2x - 1\;{\text{and }}g\left( x \right) = x^2 [/tex]

Which satisfies [itex] \forall x \ne 1,\;f\left( x \right) < g\left( x \right) [/itex]

But,
[tex] \mathop {\lim }\limits_{x \to 1} f\left( x \right) = \mathop {\lim }\limits_{x \to 1} g\left( x \right) = 1 [/tex]

and [itex] 1 =1 [/itex] :biggrin:
 
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