# L'Hospital's questions

• Apr 9th 2013, 08:31 PM
Alucard2487
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.
• Apr 9th 2013, 11:02 PM
Soroban
Re: L'Hospital's questions
Hello, Alucard2487!

Quote:

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

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

Quote:

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

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

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

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

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

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

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

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

Quote:

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

Quote:

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

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

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

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

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

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

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

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

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

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

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

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

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

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