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Thread: L'Hospital's questions

  1. #1
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    L'Hospital's questions

    I've done a bunch of practice for my quiz tomorrow. However, I can't seem to find the answers to one of the worksheets so i can check my answers.

    1. lim x->0 1-cox/x^2. My answer 1/2

    2. lim x-> infinity ln(x)^2/x. My answer 0

    3. lim x-> infinity (3x-sqrt(9x^2+2x+1)). Bit lost on this one. -sqrt(9x^2+2x+1)/(1/3x). 1/2(9x^2+2x+1)^-1/2*(18x+2)/(-3/9x^2)

    4. lim x->0+ sqrt(x)*(ln(x)). My answer 0.

    5. lim x ->infinity (3x-2/3x+1)^x No clue

    6. lim x-> infinty (e^x+2)^1/x My answer y=e^1/3

    7. lim x-> 0 arctan(x)/x^2 My answer 0

    8. lim x->0 sinh(x)/sin(x) My answer 1.
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  2. #2
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    Re: L'Hospital's questions

    Hello, Alucard2487!

    $\displaystyle 1.\;\lim_{x\to0}\frac{1-\cos x}{x^2}\quad\text{ My answer: }\tfrac{1}{2}$ . Correct!

    $\displaystyle 2.\;\lim_{x\to\infty}\frac{\ln x^2}{x}\quad\text{ My answer: } 0$ . Good!

    $\displaystyle 3.\;\lim_{x\to\infty}\left(3x-\sqrt{9x^2+2x+1}\right)$

    Multiply by $\displaystyle \tfrac{3x+\sqrt{9x^2+2x+1}}{3x+\sqrt{9x^2+2x+1}} $

    $\displaystyle \frac{3x-\sqrt{9x^2+2x+1}}{1}\cdot \frac{3x + \sqrt{9x^2+2x+1}}{3x+\sqrt{9x^2+2x+1}} \;=\;\frac{9x^2 - (9x^2 + 2x + 1)}{3x + \sqrt{9x^2+2x+1}} $

    . . . . $\displaystyle =\;\frac{-2x-1}{3x+\sqrt{9x^2+2x +1}} \qquad \text{ This goes to }\frac{\infty}{\infty}$

    Apply L'Hopital: .$\displaystyle \frac{-2}{3 + \frac{1}{2}\frac{18x+2}{\sqrt{9x^2+2x+1}}} \;=\;\frac{-2}{3 + \frac{9x+2}{\sqrt{9x^2+2x+1}}} $

    Divide numerator and denominator of that bottom fraction by $\displaystyle x:$

    . . . . $\displaystyle \frac{-2}{3 + \frac{9+\frac{2}{x}}{\sqrt{9 + \frac{2}{x} + \frac{1}{x^2}}}} $

    $\displaystyle \text{Therefore: }\:\lim_{x\to\infty} \frac{-2}{3 + \frac{9+\frac{2}{x}}{\sqrt{9 + \frac{2}{x} + \frac{1}{x^2}}}} \;=\;\frac{-2}{3+ \frac{9+0}{\sqrt{9+0+0}}} \;=\;\frac{-2}{3+\frac{9}{3}} \;=\;\frac{-2}{6} \;=\;-\frac{1}{3}$




    $\displaystyle 4.\;\lim_{x\to0^+}\sqrt x\cdot\ln x \quad\text{ My answer: }0$ . Yes!


    $\displaystyle 5.\;\lim_{x\to\infty}\left(\frac{3x-2}{3x+1}\right)^x$

    $\displaystyle \text{Let }\,y \;=\;\left(\frac{3x-2}{3x+1}\right)^x$

    $\displaystyle \text{Take logs: }\:\ln y \;=\;\ln\left(\frac{3x-2}{3x+1}\right)^x \;=\;x\ln\left(\frac{3x-2}{3x+1}\right) $

    . . . . . . . . $\displaystyle \ln y \;=\;x\big[\ln(3x-2) - \ln(3x+1)\big]$


    Differentiate: .$\displaystyle \frac{y'}{y} \;=\;x\left(\frac{3}{3x-2} - \frac{3}{3x+1}\right) + \ln\left(\frac{3x-2}{3x+1}\right) $

    . . . . . . . . . . .$\displaystyle \frac{y'}{y} \;=\;\frac{3x}{(3x-2)(3x-1)} + \ln\left(\frac{3x-2}{3x+1}\right)$


    Divide numerator and denominator of the first fraction by $\displaystyle x^2.$
    Divide numerator and denominator of the second fraction by $\displaystyle x.$

    . . . $\displaystyle \frac{\frac{3}{x}}{(3-\frac{2}{x})(3+\frac{1}{x})} + \ln\left(\frac{3-\frac{2}{x}}{3 + \frac{1}{x}}\right)$


    Hence: .$\displaystyle \lim_{x\to\infty} \left[\frac{\frac{3}{x}}{(3-\frac{2}{x})(3+\frac{1}{x})} + \ln\left(\frac{3-\frac{2}{x}}{3 + \frac{1}{x}}\right)\right]$

    . . . . . $\displaystyle =\;\frac{0}{(3-0)(3+0)} + \ln\left(\frac{3-0}{3+0}\right) $

    . . . . . $\displaystyle =\;0 + \ln1 \;=\;0+0 \;=\;0$


    We have: .$\displaystyle \lim_{x\to\infty} \ln y \:=\:0$

    Therefore: .$\displaystyle \lim_{x\to\infty}y \;=\;e^0 \;=\;1$


    It's very late here . . . *yawn*
    I'll finish up tomorrow.
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