Published on 12 Oct 2018

School

Department

Course

Professor

Name:

Exam 1 - Math 151

Sep 11, 2007

R. Ketchersid

1. [9pts] a) [3pts] Give the rigorous (“ε-δ”) deﬁnition of “limx→31

x−2= 1”.

For all ε > 0, there is a δ > 0such that

0<|x−3|< δ =⇒

1

x−2−1

< ε

b) [3pts] Find an interval (xL, xR)(for arbitrary ε) satisfying xL< x < xR→1−ε < 1

x−2<1 + ε.

(See graph.)

For xLyou must solve 1

xL−2= 1 + εso xL−2 = 1

1+εand xL= 2 + 1

1+ε. Similarly compute xR.

xL= 2 + 1

1+ε

xR= 2 + 1

1−ε

c) [3pts] Use the above to give a δεsatisfying 0<|x−3|< δε→

1

x−2−1< ε.

δε= min{3−xL, xR−3}= min 1−1

1 + ε,1

1−ε−1

= min ε

1 + ε,ε

1−ε=ε

1 + ε

Use your δεto ﬁnd: (Give actual numbers with at least 4 decimal places of accuracy.)

δ1= 0.5δ0.1= 0.0909 δ0.01 = 0.0099

-2-

On problems 2 and 3, Tmeans the fact is necessarily true, Fmeans the fact is necessarily false,

and Nmeans that there is not enough information, that is the fact could be true or false in various

cases.

2. [5pts] Suppose f(x)is continuous on [0,∞)and f(0) = 1 and limx→∞ = 0.

a) T F N f(x)has a maximum on [0,∞).

There is some x0so that for x≥x0,|f(x)|<1/2since limx→∞ f(x)=0. By extreme value theorem

f(x)has a maximum value Mon [0, x0].M≥1so Mis the maximum value of f(x)on [0,∞).

b) T F N f(x)has a minimum on [0,∞).

It is easy to imagine graphs, to come up with expressions requires a bit more thought. The

following work: f(x) = 1

x+1 clearly shows that there need not be a minimum while f(x) = 1−x

(x+1)2

has a minimum.

c) T F N There is some c > 0such that f(c)=1/2.

This is IVT. Since limx→∞ f(x)=0we have some x > 0so that f(x)<1/2. By IVT there is

c∈(0, x)with f(c) = 1/2.

d) T F N There is some c > 0such that f(c) = 0.

Same examples as in (b) work.

e) T F N There is some c > 0such that limx→cf(x)=1/3.

Just as in (c) there is some cso that f(c) = 1/3. since fis continuous at c,limx→cf(x)=1/3.

3. Problem 3 was thrown out for those students taking the version of the exam with T/F/N. The

problem should have read “g(x)is continuous at x= 4” as written it does not really make sense.

See next page for intended solution.

## Document Summary

Sep 11, 2007: ketchersid, [9pts] a) [3pts] give the rigorous ( - ) de nition of limx 3. For all > 0, there is a > 0 such that. 1 (cid:12)(cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12)(cid:12) < : [3pts] find an interval (xl, xr) (for arbitrary ) satisfying xl < x < xr 1 < 1 (see graph. ) For xl you must solve xl 2 = 1 + so xl 2 = 1. Similarly compute xr. x 2 < 1 + . 1 : [3pts] use the above to give a satisfying 0 < |x 3| < (cid:12)(cid:12)(cid:12) 1 x 2 1 (cid:26) (cid:27) = min{3 xl, xr 3} = min. 1 + (cid:12)(cid:12)(cid:12) < . (cid:27) (cid:26) Use your to nd: (give actual numbers with at least 4 decimal places of accuracy. ) There is some x0 so that for x x0, |f(x)| < 1/2 since limx f(x) = 0. By extreme value theorem f(x) has a maximum value m on [0, x0].