Results 1 to 5 of 5
Like Tree4Thanks
  • 1 Post By Prove It
  • 1 Post By kalyanram
  • 1 Post By Prove It
  • 1 Post By Prove It

Math Help - Limit Problems

  1. #1
    Newbie
    Joined
    Aug 2012
    From
    Chile
    Posts
    1

    Limit Problems

    Hi, I'm stuck on these problems, help is appreciated. Thank you.

    1. Is there an a\in \mathbb{R} for which \mathop {\lim }\limits_{x \to  - 2} {{3{x^2} + ax + a + 3} \over {{x^2} + x - 2}} exists?

    2. \underset{x\to 1}{\mathop{\lim }}\,\frac{{{x}^{m}}-1}{{{x}^{n}}-1} \text{   }m\text{, }n\text{ }\in \text{ }\mathbb{Z^{+}}

    3. \underset{x\to \infty }{\mathop{\lim }}\,\frac{\sqrt{x}-1}{x+1}
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,548
    Thanks
    1418

    Re: Limit Problems

    Quote Originally Posted by Grundsatz View Post
    Hi, I'm stuck on these problems, help is appreciated. Thank you.

    1. Is there an a\in \mathbb{R} for which \mathop {\lim }\limits_{x \to  - 2} {{3{x^2} + ax + a + 3} \over {{x^2} + x - 2}} exists?

    2. \underset{x\to 1}{\mathop{\lim }}\,\frac{{{x}^{m}}-1}{{{x}^{n}}-1} \text{   }m\text{, }n\text{ }\in \text{ }\mathbb{Z^{+}}

    3. \underset{x\to \infty }{\mathop{\lim }}\,\frac{\sqrt{x}-1}{x+1}
    For 1. the denominator is \displaystyle \begin{align*} x^2 + x - 2 = (x + 2)(x - 1) \end{align*}. To be able to evaluate the limit, you need to be able to cancel \displaystyle \begin{align*} (x + 2) \end{align*} in the numerator. So \displaystyle \begin{align*} (x + 2) \end{align*} has to be a factor.

    Using the factor theorem, \displaystyle \begin{align*} P(x) = 3x^2 + a\,x + a + 3 \end{align*} will have \displaystyle \begin{align*} (x + 2) \end{align*} as a factor if \displaystyle \begin{align*} P(-2) = 0 \end{align*}. So we have

    \displaystyle \begin{align*} P(-2) &= 0 \\ 3(-2)^2 + a(-2) + a + 3 &= 0 \\ 12 - 2a + a + 3 &= 0 \\ 15 - a &= 0 \\ a &= 15 \end{align*}

    So in order for \displaystyle \begin{align*} \lim_{x \to -2} \frac{3x^2 + a\,x + a + 3}{x^2 + x - 2} \end{align*} to exist, \displaystyle \begin{align*} a = 15 \end{align*}.
    Thanks from Grundsatz
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member kalyanram's Avatar
    Joined
    Jun 2008
    From
    Bangalore, India
    Posts
    143
    Thanks
    14

    Re: Limit Problems

    Quote Originally Posted by Grundsatz View Post
    1. Is there an a\in \mathbb{R} for which \mathop {\lim }\limits_{x \to  - 2} {{3{x^2} + ax + a + 3} \over {{x^2} + x - 2}} exists?
    Hint: x^2+x-2 = (x+2)(x-1) so can you get rid of (x+2). How can you do that?

    Quote Originally Posted by Grundsatz View Post
    2. \underset{x\to 1}{\mathop{\lim }}\,\frac{{{x}^{m}}-1}{{{x}^{n}}-1} \text{   }m\text{, }n\text{ }\in \text{ }\mathbb{Z^{+}}
    Try factorizing x^m-1, x^n-1 do they have a common factor?

    Quote Originally Posted by Grundsatz View Post
    3. \underset{x\to \infty }{\mathop{\lim }}\,\frac{\sqrt{x}-1}{x+1}
    This one is easy if you get the first two.
    Thanks from Grundsatz
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,548
    Thanks
    1418

    Re: Limit Problems

    Quote Originally Posted by Grundsatz View Post
    Hi, I'm stuck on these problems, help is appreciated. Thank you.

    1. Is there an a\in \mathbb{R} for which \mathop {\lim }\limits_{x \to  - 2} {{3{x^2} + ax + a + 3} \over {{x^2} + x - 2}} exists?

    2. \underset{x\to 1}{\mathop{\lim }}\,\frac{{{x}^{m}}-1}{{{x}^{n}}-1} \text{   }m\text{, }n\text{ }\in \text{ }\mathbb{Z^{+}}

    3. \underset{x\to \infty }{\mathop{\lim }}\,\frac{\sqrt{x}-1}{x+1}
    \displaystyle \begin{align*} \lim_{x \to 1}{\frac{x^m - 1}{x^n - 1}} &= \lim_{x \to 1}{\frac{(x - 1)\left( x^{m-1} + x^{m - 2} + x^{m - 3} + \dots + 1  \right)}{(x - 1)\left( x^{n - 1} + x^{n - 2} + x^{n - 3} + \dots + 1 \right)}} \\ &= \lim_{x \to 1}{\frac{x^{m-1} + x^{m-2} + x^{m-3} + \dots + 1}{x^{n - 1} + x^{n-2} + x^{n-3} + \dots + 1}} \\ &= \frac{1 + 1 + 1 + \dots + 1\,\,\, \textrm{(}m \textrm{ times)}}{1 + 1 + 1 + \dots + 1 \,\,\, \textrm{(}n \textrm{ times)}} \\ &= \frac{m}{n} \end{align*}
    Thanks from Grundsatz
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,548
    Thanks
    1418

    Re: Limit Problems

    Quote Originally Posted by Grundsatz View Post
    Hi, I'm stuck on these problems, help is appreciated. Thank you.

    1. Is there an a\in \mathbb{R} for which \mathop {\lim }\limits_{x \to  - 2} {{3{x^2} + ax + a + 3} \over {{x^2} + x - 2}} exists?

    2. \underset{x\to 1}{\mathop{\lim }}\,\frac{{{x}^{m}}-1}{{{x}^{n}}-1} \text{   }m\text{, }n\text{ }\in \text{ }\mathbb{Z^{+}}

    3. \underset{x\to \infty }{\mathop{\lim }}\,\frac{\sqrt{x}-1}{x+1}
    3. Rationalising the numerator gives

    \displaystyle \begin{align*} \lim_{x \to \infty}{\frac{\sqrt{x} - 1}{x + 1}} &= \lim_{x\to \infty}{\frac{\left( \sqrt{x} - 1 \right)\left( \sqrt{x} + 1 \right)}{(x + 1)\left( \sqrt{x} + 1 \right)}} \\ &= \lim_{x \to \infty}{\frac{x - 1}{(x + 1)\left(\sqrt{x} + 1 \right)}} \\ &= \lim_{x \to \infty}{\frac{x + 1 - 2}{(x + 1)\left( \sqrt{x} + 1 \right)}} \\ &= \lim_{x \to \infty}{\left[ \frac{1}{\sqrt{x} + 1} - \frac{2}{(x + 1)\left( \sqrt{x} + 1 \right)} \right]} \\ &= \frac{1}{\sqrt{1} + 1} - \frac{2}{(1 + 1)\left( \sqrt{1} + 1 \right)} \\ &= \frac{1}{2} - \frac{1}{2} \\ &= 0 \end{align*}
    Thanks from Grundsatz
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. 2 limit problems i need help with
    Posted in the Calculus Forum
    Replies: 3
    Last Post: February 12th 2010, 01:39 PM
  2. Limit Problems Help
    Posted in the Calculus Forum
    Replies: 6
    Last Post: October 20th 2008, 01:05 PM
  3. 4 Limit problems.
    Posted in the Calculus Forum
    Replies: 11
    Last Post: October 2nd 2008, 09:33 AM
  4. limit problems
    Posted in the Calculus Forum
    Replies: 5
    Last Post: May 6th 2008, 08:19 PM
  5. help with two limit problems please!
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 30th 2008, 12:57 AM

Search Tags


/mathhelpforum @mathhelpforum