- #1
FarazAli
- 16
- 0
The rigorous definition given in my calculus book is as follows
[tex]\lim_{x\to{a}}f(x) = L[/tex] if given any number [tex]\epsilon > 0[/tex] we can find a number [tex]\delta > 0[/tex] such that [tex]| f(x) - L | < \epsilon[/tex] if [tex] 0 < | x - a | < \delta[/tex]
How can I use that to prove [tex]\lim_{x\to{49}} x^{\frac{1}{2}} = 7[/tex]
This is what I did
[tex]| f(x) - L | < \epsilon[/tex]
[tex]| x^{\frac{1}{2}} - 7 | < \epsilon[/tex]
[tex]| x - 7x^{\frac{1}{2}} + 49 | < \epsilon^2[/tex]
That's where I get stuck and don't know where to go. This is the first time I've encountered any epsilons and deltas so its all confusing. Thanks for any help
[tex]\lim_{x\to{a}}f(x) = L[/tex] if given any number [tex]\epsilon > 0[/tex] we can find a number [tex]\delta > 0[/tex] such that [tex]| f(x) - L | < \epsilon[/tex] if [tex] 0 < | x - a | < \delta[/tex]
How can I use that to prove [tex]\lim_{x\to{49}} x^{\frac{1}{2}} = 7[/tex]
This is what I did
[tex]| f(x) - L | < \epsilon[/tex]
[tex]| x^{\frac{1}{2}} - 7 | < \epsilon[/tex]
[tex]| x - 7x^{\frac{1}{2}} + 49 | < \epsilon^2[/tex]
That's where I get stuck and don't know where to go. This is the first time I've encountered any epsilons and deltas so its all confusing. Thanks for any help