Proving function limit using $epsilon$-$delta$ definition
$begingroup$
Guess the following limits and prove your answers using the epsilon-delta definition:
$ 1) limlimits_{xto 2} (x^2-2x)$
$ 2) limlimits_{xto2} dfrac{5x+1}{2x-5} $
I understand the definition, but having a problem putting it into work:
1) The guess is obvious, $ lim = 0 $:
$$ lvert x^2-2x-0 rvert = lvert x(x-2) rvert = lvert x rvert cdot lvert x-2 rvert $$
Can I finish this with saying:
$$ lvert x rvert cdot lvert x-2 rvert leq lvert x-2 rvert$$
And then picked $ delta = epsilon $ and I'm done??
2) The guess is $ lim = -11 $.
$$ biggrlvert {5x+1 over 2x-5} - ( -11 ) biggrrvert = biggrlvert {27x-54 over 2x-5} biggrrvert = biggrlvert {27(x-2) over 2x-5} biggrlvert leq lvert 27(x-2) rvert = 27lvert x- 2rvert $$
But I'm quite lost how to continue from here..
Seems like I just need the final punch on this, thanks in advance.
calculus limits
edited Dec 17 '18 at 16:14
Tianlalu
3,09421138
asked Dec 17 '18 at 16:11
TegernakoTegernako
908
$endgroup$
add a comment |
$begingroup$
Guess the following limits and prove your answers using the epsilon-delta definition:
$ 1) limlimits_{xto 2} (x^2-2x)$
$ 2) limlimits_{xto2} dfrac{5x+1}{2x-5} $
I understand the definition, but having a problem putting it into work:
1) The guess is obvious, $ lim = 0 $:
$$ lvert x^2-2x-0 rvert = lvert x(x-2) rvert = lvert x rvert cdot lvert x-2 rvert $$
Can I finish this with saying:
$$ lvert x rvert cdot lvert x-2 rvert leq lvert x-2 rvert$$
And then picked $ delta = epsilon $ and I'm done??
2) The guess is $ lim = -11 $.
$$ biggrlvert {5x+1 over 2x-5} - ( -11 ) biggrrvert = biggrlvert {27x-54 over 2x-5} biggrrvert = biggrlvert {27(x-2) over 2x-5} biggrlvert leq lvert 27(x-2) rvert = 27lvert x- 2rvert $$
But I'm quite lost how to continue from here..
Seems like I just need the final punch on this, thanks in advance.
calculus limits
edited Dec 17 '18 at 16:14
Tianlalu
3,09421138
asked Dec 17 '18 at 16:11
TegernakoTegernako
908
$endgroup$
add a comment |
$begingroup$
Guess the following limits and prove your answers using the epsilon-delta definition:
$ 1) limlimits_{xto 2} (x^2-2x)$
$ 2) limlimits_{xto2} dfrac{5x+1}{2x-5} $
I understand the definition, but having a problem putting it into work:
1) The guess is obvious, $ lim = 0 $:
$$ lvert x^2-2x-0 rvert = lvert x(x-2) rvert = lvert x rvert cdot lvert x-2 rvert $$
Can I finish this with saying:
$$ lvert x rvert cdot lvert x-2 rvert leq lvert x-2 rvert$$
And then picked $ delta = epsilon $ and I'm done??
2) The guess is $ lim = -11 $.
$$ biggrlvert {5x+1 over 2x-5} - ( -11 ) biggrrvert = biggrlvert {27x-54 over 2x-5} biggrrvert = biggrlvert {27(x-2) over 2x-5} biggrlvert leq lvert 27(x-2) rvert = 27lvert x- 2rvert $$
But I'm quite lost how to continue from here..
Seems like I just need the final punch on this, thanks in advance.
calculus limits
edited Dec 17 '18 at 16:14
Tianlalu
3,09421138
asked Dec 17 '18 at 16:11
TegernakoTegernako
908
$endgroup$
Guess the following limits and prove your answers using the epsilon-delta definition:
$ 1) limlimits_{xto 2} (x^2-2x)$
$ 2) limlimits_{xto2} dfrac{5x+1}{2x-5} $
I understand the definition, but having a problem putting it into work:
1) The guess is obvious, $ lim = 0 $:
$$ lvert x^2-2x-0 rvert = lvert x(x-2) rvert = lvert x rvert cdot lvert x-2 rvert $$
Can I finish this with saying:
$$ lvert x rvert cdot lvert x-2 rvert leq lvert x-2 rvert$$
And then picked $ delta = epsilon $ and I'm done??
2) The guess is $ lim = -11 $.
$$ biggrlvert {5x+1 over 2x-5} - ( -11 ) biggrrvert = biggrlvert {27x-54 over 2x-5} biggrrvert = biggrlvert {27(x-2) over 2x-5} biggrlvert leq lvert 27(x-2) rvert = 27lvert x- 2rvert $$
But I'm quite lost how to continue from here..
Seems like I just need the final punch on this, thanks in advance.
calculus limits
calculus limits
edited Dec 17 '18 at 16:14
Tianlalu
3,09421138
asked Dec 17 '18 at 16:11
TegernakoTegernako
908
edited Dec 17 '18 at 16:14
Tianlalu
3,09421138
asked Dec 17 '18 at 16:11
TegernakoTegernako
908
edited Dec 17 '18 at 16:14
Tianlalu
3,09421138
edited Dec 17 '18 at 16:14
Tianlalu
3,09421138
edited Dec 17 '18 at 16:14
Tianlalu
3,09421138
3,09421138
asked Dec 17 '18 at 16:11
TegernakoTegernako
908
asked Dec 17 '18 at 16:11
TegernakoTegernako
908
asked Dec 17 '18 at 16:11
TegernakoTegernako
908
908
add a comment |
add a comment |
1 Answer
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$begingroup$
For 1), you must take a neighborhood of 2 (let says $[1,3]$), and then, $$|x||x-2|leq 3|x-2|.$$ Then, $delta =min{1, frac{varepsilon }{3}}$ work. For 2), take $delta =frac{varepsilon }{27}$.
answered Dec 17 '18 at 16:18
NewMathNewMath
4059
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
For 1), you must take a neighborhood of 2 (let says $[1,3]$), and then, $$|x||x-2|leq 3|x-2|.$$ Then, $delta =min{1, frac{varepsilon }{3}}$ work. For 2), take $delta =frac{varepsilon }{27}$.
answered Dec 17 '18 at 16:18
NewMathNewMath
4059
$endgroup$
add a comment |
$begingroup$
For 1), you must take a neighborhood of 2 (let says $[1,3]$), and then, $$|x||x-2|leq 3|x-2|.$$ Then, $delta =min{1, frac{varepsilon }{3}}$ work. For 2), take $delta =frac{varepsilon }{27}$.
answered Dec 17 '18 at 16:18
NewMathNewMath
4059
$endgroup$
add a comment |
$begingroup$
For 1), you must take a neighborhood of 2 (let says $[1,3]$), and then, $$|x||x-2|leq 3|x-2|.$$ Then, $delta =min{1, frac{varepsilon }{3}}$ work. For 2), take $delta =frac{varepsilon }{27}$.
answered Dec 17 '18 at 16:18
NewMathNewMath
4059
$endgroup$
For 1), you must take a neighborhood of 2 (let says $[1,3]$), and then, $$|x||x-2|leq 3|x-2|.$$ Then, $delta =min{1, frac{varepsilon }{3}}$ work. For 2), take $delta =frac{varepsilon }{27}$.
answered Dec 17 '18 at 16:18
NewMathNewMath
4059
edited Dec 17 '18 at 16:27
answered Dec 17 '18 at 16:18
NewMathNewMath
4059
answered Dec 17 '18 at 16:18
NewMathNewMath
4059
answered Dec 17 '18 at 16:18
NewMathNewMath
4059
4059
add a comment |
add a comment |
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