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Delta-epsilon proofs (again)

  • Thread starter drawar
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Homework Statement


Prove the following limits using delta-epsilon definition
(a)[itex]\mathop {\lim }\limits_{x \to 5{}^ - } \sqrt[4]{{5 - x}} = 0[/itex]
(b)[itex]\mathop {\lim }\limits_{x \to 2} ({x^4} - 1) = 15[/itex]
(c)[itex]\mathop {\lim }\limits_{x \to - 1} \frac{{{x^2} - 2x}}{{x + 2}} = 3[/itex]


Homework Equations





The Attempt at a Solution


So here's my working, please let me know if there's any correction needed. Thank you!

(a) Given ε>0, choose δ=ε^4. Then:
[itex]0 < 5 - x < \delta \Rightarrow \left| {\sqrt[4]{{5 - x}}} \right| = \sqrt[4]{{5 - x}} < \sqrt[4]{\delta } = \sqrt[4]{{{\varepsilon ^4}}} = \varepsilon [/itex]

(b) Given ε>0, choose δ=min{1; ε/65}, then:
[itex]0 < \left| {x - 2} \right| < \delta \Rightarrow \left| {({x^4} - 1) - 15} \right| = \left| {{x^4} - 16} \right| = ({x^2} + 4)\left| {x + 2} \right|\left| {x - 2} \right| = \left[ {{{(x - 2)}^2} + 4(x - 2) + 8} \right]\left| {x - 2 + 4} \right|\left| {x - 2} \right| \le \left( {{{\left| {x - 2} \right|}^2} + 4\left| {x - 2} \right| + 8} \right)\left( {\left| {x - 2} \right| + 4} \right)\left| {x - 2} \right| < \left( {{\delta ^2} + 4\delta + 8} \right)(\delta + 4)\delta < (13)(5)\delta < (65)\frac{\varepsilon }{{65}} = \varepsilon [/itex]

(c) Given ε>0, choose δ=min{1/2;ε/15}. Then:
[itex]0 < \left| {x + 1} \right| < \delta \Rightarrow \left| {\frac{{{x^2} - 2x}}{{x + 2}} - 3} \right| = \left| {\frac{{{x^2} - 5x - 6}}{{x + 2}}} \right| = \left| {\frac{{{{(x + 1)}^2} - 7(x + 1)}}{{x + 2}}} \right| \le \frac{1}{{\left| {x + 2} \right|}}\left( {{{\left| {x + 1} \right|}^2} + 7\left| {x + 1} \right|} \right)[/itex]
But [itex]\left| {x + 1} \right| < \frac{1}{2} \Leftrightarrow \frac{1}{2} < \left| {x + 2} \right| < \frac{3}{2}[/itex]
Hence
[itex]\frac{1}{{\left| {x + 2} \right|}}\left( {{{\left| {x + 1} \right|}^2} + 7\left| {x + 1} \right|} \right) < 2({\delta ^2} + 7\delta ) = 2\delta (\delta + 7) < (2)(\frac{\varepsilon }{{15}})(\frac{{15}}{2}) = \varepsilon [/itex]
 

Answers and Replies

  • #2
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Seems OK.
 
  • #3
SammyS
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Homework Statement


Prove the following limits using delta-epsilon definition
(a)[itex]\mathop {\lim }\limits_{x \to 5{}^ - } \sqrt[4]{{5 - x}} = 0[/itex]
(b)[itex]\mathop {\lim }\limits_{x \to 2} ({x^4} - 1) = 15[/itex]
(c)[itex]\mathop {\lim }\limits_{x \to - 1} \frac{{{x^2} - 2x}}{{x + 2}} = 3[/itex]

The Attempt at a Solution


So here's my working, please let me know if there's any correction needed. Thank you!

(b) Given ε>0, choose δ=min{1; ε/65}, then:
[itex]0 < \left| {x - 2} \right| < \delta \Rightarrow \left| {({x^4} - 1) - 15} \right| = \left| {{x^4} - 16} \right| = ({x^2} + 4)\left| {x + 2} \right|\left| {x - 2} \right| [/itex]
[itex]= \left[ {{{(x - 2)}^2} + 4(x - 2) + 8} \right]\left| {x - 2 + 4} \right|\left| {x - 2} \right| \le \left( {{{\left| {x - 2} \right|}^2} + 4\left| {x - 2} \right| + 8} \right)\left( {\left| {x - 2} \right| + 4} \right)\left| {x - 2} \right| < \left( {{\delta ^2} + 4\delta + 8} \right)(\delta + 4)\delta < (13)(5)\delta < (65)\frac{\varepsilon }{{65}} = \varepsilon [/itex]​
Looks good, but it does leave the reader with a lot of work to do, to verify where all that algebra came from & where it's going.

Also, break up that overly long LaTeX line.

[itex]0 < \left| {x - 2} \right| < \delta \Rightarrow \left| {({x^4} - 1) - 15} \right| = \left| {{x^4} - 16} \right| = ({x^2} + 4)\left| {x + 2} \right|\left| {x - 2} \right| [/itex]
[itex]= \left[ {{{(x - 2)}^2} + 4(x - 2) + 8} \right]\left| {x - 2 + 4} \right|\left| {x - 2} \right| \le \left( {{{\left| {x - 2} \right|}^2} + 4\left| {x - 2} \right| + 8} \right)\left( {\left| {x - 2} \right| + 4} \right)\left| {x - 2} \right| [/itex]
[itex]< \left( {{\delta ^2} + 4\delta + 8} \right)(\delta + 4)\delta < (13)(5)\delta < (65)\frac{\varepsilon }{{65}} = \varepsilon [/itex]​
 
Last edited:

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