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October 29th, 2012, 12:28 PM   #1
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Calculus, limits, four problems

Hello to everyone! I have four problem I can't solve.
Please, help me. Thanks a lot beforehand
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 problem11.png (1.1 KB, 172 views) problem12.png (912 Bytes, 172 views) problem13.png (1,009 Bytes, 172 views) problem14.png (1.1 KB, 172 views)

 October 29th, 2012, 12:34 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Calculus, limits, four problems First, are you "allowed" to use L'Hôpital's rule?
 October 29th, 2012, 12:57 PM #3 Newbie   Joined: Dec 2011 Posts: 13 Thanks: 0 Re: Calculus, limits, four problems I don't know about L'Hôpital's rule yet. I just have started to learn calculus. For examples of solutions I try to insight in understanding of limits. So, I ask of you to solve this limits and try to explain me how you do it.
 October 29th, 2012, 01:07 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Calculus, limits, four problems Okay, I just wanted to know first before working solutions, only to be told afterwards that L'Hôpital's rule is not allowed. It happens quite a bit with these types of limits.
 October 29th, 2012, 11:50 PM #5 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Calculus, limits, four problems I'll work the first one: 1.) $\lim_{x\to\infty}\frac{x^2-\sqrt{3x^5-7}}{$$x^2-x\cos(x)+1$$\sqrt{x}}=L$ I would break this up into two limits: $L_1=\lim_{x\to\infty}\frac{x^2}{$$x^2-x\cos(x)+1$$\sqrt{x}}$ Here, the degree in the denominator is greater than that of the numerator, hence: $L_1=0$ $L_2=\lim_{x\to\infty}\frac{\sqrt{3x^5-7}}{$$x^2-x\cos(x)+1$$\sqrt{x}}$ $L_2=\sqrt{\lim_{x\to\infty}\frac{3x^5-7}{x$$x^2-x\cos(x)+1$$^2}}$ The degree in the numerator and denominator is the same, so we take the ratio of the leading coefficients: $L_2=\sqrt{3}$ And so we have: $L=L_1-L_2=-\sqrt{3}$
 October 30th, 2012, 03:47 AM #6 Math Team     Joined: Aug 2012 From: Sana'a , Yemen Posts: 1,177 Thanks: 44 Math Focus: Theory of analytic functions Re: Calculus, limits, four problems If you mean by tg(x) tan(x) then the third and fourth one can be solved by multiplying and dividing by the conjugate.
October 30th, 2012, 07:42 AM   #7
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Re: Calculus, limits, four problems

Hello, The Crisp!

I'll do the last one . . .

Quote:
 $\lim_{x\to\infty}\,\left(\sqrt{x^2\,-\,1}\,-\,\sqrt{x^2\,+\,1\right)$

Multiply numerator and denominator by the conjugate:

[color=beige]. . [/color]$\frac{\sqrt{x^2\,-\,1}\,-\,\sqrt{x^2\,+\,1}}{1}\,\cdot\,\frac{\sqrt{x^2\,-\,1}\,+\,\sqrt{x^2\,+\,1}}{\sqrt{x^2\,-\,1}\,+\,\sqrt{x^2\,+\,1}} \;=\;\frac{(x^2\,-\,1)\,-\,(x^2\,+\,1)}{\sqrt{x^2\,-\,1}\,+\,\sqrt{x^2\,+\,1}} \;=\;\frac{-2}{\sqrt{x^2\,-\,1}\,+\,\sqrt{x^2\,+\,1}}$

Divide numerator and denominator by $x:$

[color=beige]. . [/color]$\frac{-\frac{2}{x}}{\frac{\sqrt{x^2-1}}{x}\,+\,\frac{\sqrt{x^2+1}}{x}} \;=\;\frac{-\frac{2}{x}}{\sqrt{\frac{x^2-1}{x^2}}\,+\,\sqrt{\frac{x^2+1}{x^2}}} \;=\;\frac{-\frac{2}{x}}{\sqrt{1-\frac{1}{x^2}}\,+\,\sqrt{1+\frac{1}{x^2}}}$

Therefore:[color=beige] .[/color]$\lim_{x\to\infty}\,\frac{-\frac{2}{x}}{\sqrt{1-\frac{1}{x^2}}\,+\,\sqrt{1+\frac{1}{x^2}}} \;=\;\frac{0}{\sqrt{1-0}\,+\,\sqrt{1+0}} \;=\;\frac{0}{2} \;=\;0$

 October 30th, 2012, 09:08 AM #8 Newbie   Joined: Dec 2011 Posts: 13 Thanks: 0 Re: Calculus, limits, four problems Thanks to everyone for answers! I try to solve problem#12. Please, check it ! Problem#12
 October 30th, 2012, 09:10 AM #9 Newbie   Joined: Dec 2011 Posts: 13 Thanks: 0 Re: Calculus, limits, four problems P.S. Sorry, Iforgot to rotate the picture
October 31st, 2012, 11:16 AM   #10
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From: Sana'a , Yemen

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Math Focus: Theory of analytic functions
Re: Calculus, limits, four problems

Quote:
 Originally Posted by The Crisp P.S. Sorry, Iforgot to rotate the picture
I think you have got the idea.

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