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Some calculus problems: limits, derivatives, and simplification.

  1. Dec 18, 2007 #1
    I'm going to post the question and the work i have done thus far. I just need to know if I am on the right track with the problems and how I can go about simplifying them to finish them off.

    These are from a class homework assignment that will be graded.

    lim (sqrt((9x^2)+4))/x+3

    lim (1/2)((9x^2)+4)^(-1/2)(18x) //L'Hopital rule

    from there i got

    lim (9x)/(sqrt((9x^2)+4))

    where do i go from here?

    lim sqrt(x^2 +6)/4x+7

    lim ((.5)(x^2 +6)^-1/2 (2x)) / 4

    where to from here?

    lim sinxcosx
    x-> (2pi)/3

    i got that it equals -.433

    for these i need to find the dy/dx


    i got

    y'= (((6x+3)^(-1/2))[(1/2)(4x-5)-12(6x+3)]) / ((4x-5)^4)


    i got

    y'= (((5x+2)^-.5)[(5/2)(3x-1)-15])/((3x-1)^6)

    y= (x^4)cotx

    i got

    y'= (-x^4)cscx + (4x^3)cotx


    i got

    y'=(-3x^2)cscxcotx + 6xcscx

    some other weird problems:

    find f'(3) using the alternate form of a derivative if f(x) is | x-3 |

    well.....all i know is f'(c)= lim as x -> c is (f(x)-f(c))/x-c

    but do i need to break it down into piece-wise stuff to solve this one?

    y=sin3x on [0, pi/6]


    this is mean value theorem so, i solved for both endpoints and got the dy/dx of the function and set it equal to the slope of the secant line...im suck on the step where i set it equal...so i have:

    (3(sqrt(3)))= cos(2x)

    i believe i need to use cos double angle theorem or something but i'm not sure how that would work.

    any help on these would be GREATLY appreciated, this hw is not due until friday so I'm not last minuting it or anything...been working on a giant packet of this stuff for 3 weeks UGH...its all fun though :D
  2. jcsd
  3. Dec 18, 2007 #2
    ok i don't want to make any mistakes in interpreting your problem, so first one

    [tex]\lim_{x \rightarrow \infty}\frac{\sqrt{9x^2 +4}}{x+3}[/tex]


    1. Rather than using L'Hopital's rule, can you simplify this algebraically?
    Last edited: Dec 18, 2007
  4. Dec 18, 2007 #3
    is that is the correct problem...i have no clue how to simplify it algebraically...
  5. Dec 18, 2007 #4
    The trick is to divide both the numerator and the denominator by a power of x
    Watch this

    [tex]\lim_{x\to\infty} \frac{\sqrt{9x^2+4} / x}{(x+3)/x}[/tex]

    [tex]\lim_{x\to\infty} \frac{\sqrt{9+\frac{4}{x^2}}}{1+\frac{3}{x}}[/tex]

    now can you see what to do?
  6. Dec 19, 2007 #5

    Gib Z

    User Avatar
    Homework Helper

    Or expand the numerator with Binomial series n=1/2

    [tex]\sqrt{9x^2 +4} = 3x + O(x^{1/2})[/tex].
  7. Dec 19, 2007 #6
    can i just do something like this
    sqrt(9x^2 + 4) -> 3x+2
  8. Dec 19, 2007 #7
    edit: i got 3 as the answer
  9. Dec 19, 2007 #8
    So you're saying if you take the square root of

    [tex]\sqrt{9x^2 +4}[/tex]

    You can get


    Exactly how is that allowed? If it didn't have +4, you could.
  10. Dec 19, 2007 #9
  11. Dec 19, 2007 #10
    i don't understand how that works. how can he just change stuff inside the radical
  12. Dec 19, 2007 #11
    Well first we know that, when we manipulate a problem, we have to do it in a sense that it remains the same problem: multiplication of 1, addition of 0 ... etc.



    Now I'm going to manipulate it by multiplying & dividing by x/x


    Notice: [tex]x=\sqrt{x^2}=x^{2\times\frac{1}{2}}=x[/tex]

    Now it has the same exponent of 1/2, I can combine the 2 and write it under one radical

    1) [tex]\sqrt{a}\times\sqrt{b}=a^{\frac{1}{2}}\times b^{\frac{1}{2}}=(ab)^{\frac{1}{2}}[/tex]

    2) [tex]\frac{\sqrt{a}}{\sqrt{b}}=\frac{a^{\frac{1}{2}}}{b^{\frac{1}{2}}}=(\frac{a}b})^{\frac{1}{2}}[/tex]

    So ...

    [tex]\sqrt{x}\times\frac{\sqrt{x^2}}{\sqrt{x^2}} \rightarrow \frac{\sqrt{x^3}}{\sqrt{x^2}}[/tex]

    No getting my original problem back

    [tex]\frac{\sqrt{x^3}}{\sqrt{x^2}} \rightarrow \sqrt{\frac{x^3}{x^2}} \rightarrow \sqrt{x^{3-2}}=\sqrt{x}[/tex]
    Last edited: Dec 19, 2007
  13. Dec 19, 2007 #12
    Chicken used a clever form of one....

    What happens as x approaches infinity of 1/x? What can we call this?
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