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Check my CALCULUS work, please! :)

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Question 1.
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! :) - ,

    #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! :) - ,

    #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

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