# Find the limit L. Then use the ε-δ definition to prove that the limit is L.?

lim √x

x->9

### 1 Answer

- Mr.PersonaLv 54 weeks ago
I'll take a stab:

Since sqrt(x) is only defined on the nonnegative real numbers, it follows that |sqrt(x) + 3| = sqrt(x) + 3 >= 0 + 3 = 3. I.e. 1/|sqrt(x) + 3| < 1/3. Pick δ = min {4, 3ε}. With regards to the prior, suppose not, you'd find that we'd consider |x - 3| < 4, i.e. -4 < x - 3 < 4, i.e. -1 < x < 7, i.e. x < -1 which would yield an undefined sqrt(x).

It follows that |sqrt(x) - 3| = |sqrt(x) - 3||sqrt(x) + 3|/|sqrt(x) + 3| = |x - 9|/|sqrt(x) + 3| < δ/|sqrt(x) + 3| < δ/3 = 3ε/3 = ε.

Also, learn to recognize similar problems, that way, if you do need help, you don't have to await a response from a person. E.g.

https://math.stackexchange.com/questions/310611/ep...

My hope is that you would've known that |x - 9| = |sqrt(x) - 3||sqrt(x) + 3|, but then had issues writing the delta since just dividing would yield delta = epsilon/|sqrt(x) + 3| which is dependent on x. At this point though, you should analyze if there is a bound on the term dependent on x. Thus, you'd find that it's bound by 3, i.e. the reciprocal is bound by 1/3. As for recognizing the issue of x < 0 and sqrt(x) being undefined, I wouldn't necessarily have expected you to determine how you should formulate your proof in solving, but to recognize some issue must arise since sqrt(x) isn't defined on all of R, so something bad would happen for large enough delta.

- Login to reply the answers