Sunday

May 1, 2016
Posted by **Samantha** on Monday, November 11, 2013 at 9:24am.

lim h->0(sqrt 49+h-7)/h =

14

1/14***

0

7

-1/7

Question 2.

lim x->infinity(12+x-3x^2)/(x^2-4)=

-3***

-2

0

2

3

Question 3.

lim x->infinity (5x^3+x^7)/(e^x)=

infinity***

0

-1

3

Question 4. Given that:

x 6.8 6.9 6.99 7.01 7.1 7.2

g(x) 9.44 10.21 10.92 -11.08 -11.31 -12.56

it would appear that lim x->7 g(x)=

0

7

11***

The limit does not exist.

x + 4

Question 5. Let f be defined as follows, where a ≠ 0,

f(x)={(x^2-2ax+a^2)/(x-a) if x ≠ a

{5 if x=a

Which of the following are true about f ?

I. lim f(x) x-> a exists.

II. f(a) exists.

III. f(x) is continuous at x=a.

None

I, II, and III

I only

II only

I and II only***

Question 7. lim x-> infinity (2+x-x^2)/(2x+sqrt(4x^4-3)) =

infinity

0

1***

4

-1/2

Question 8. lim x->0+ (cosx/x)=

0

1/2

1

sqrt(2)/2

The limit does not exist.***

Question 9. If g(x) is continuous for all real numbers and g(3) = −1, g(4) = 2, which of the following are necessarily true?

I. g(x)=1 at least once.

II. lim x->3.5 g(x)=g(3.5)

III. lim x->3- g(x) = lim x->3+ g(x)

I only

II only

I and II only

I, II, and III

None of these

Question 10. If the following function is continuous, what is the value of a + b?

h(t){3t^2-2t + 1, if t<0

{acos(t)+b, if 0≤t≤pi/3

{4sin^2t, if t>pi/3

0

1

2

3

4

- Check my CALCULUS work, please! :) -
**Reiny**, Monday, November 11, 2013 at 12:18pm#1 and #2 are correct

#3 --- here is a neat trick that works for most limit questions.

Use your calculator and try a number close to your approach value

e.g. for #1 I used h = .001 and then evaluated the expression

for #3, try a "large" number. However for this one even a good calculator overloads even for relative small "large" numbers

I tried x = 1000 and got ERROR 1 on my calculator

I backed right down to x = 50 , and the result was

appr 1.5 x 10^-10

looks very close to zero to me.

#4 all values <7 yield positive number getting close to 11 and values >7 are suddenly negative but close to -11

I would say the limit does not exist.

skipping #5

#7 intuitive approach ...

as x --> infinity, √(4x^4 - 3) ---> 2x^

so we approach (2 + x - x^2)/(2x + 2x^2

which approaching -1/2

(also try my method of checking with a calculator)

I used x = 1000 and got -.499 or close to -1/2

#8 correct - Check my CALCULUS work, please! :) -
**Steve**, Monday, November 11, 2013 at 12:23pm#3 is 0. exponentials grow much faster than any power of x.

l'Hospital's Rule shows that given enough iterations, the derivatives in the numerator go to zero while that pesky e^x remains in the bottom.

#4. Note that g(x) changes sign, so the limit does not exist. The limit from the left is different from the limit on the right.

#7. As x gets huge, only the highest power matters, so

(2+x-x^2)/(2x+sqrt(4x^4-3)) -> -x^2/√4x^4 = -x^2/2x^2 = -1/2

#9. all are true

#10. h must be continuous at x=0, so since

lim x->0- = 1

acos(0)+b = a+b

we must have

1 = a+b