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. $\underset{x\to \infty }{\mathop{\lim }}\,\frac{\sqrt{x}-1}{x+1}$

2. ## Re: Limit Problems

Originally Posted by Grundsatz
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*}.

3. ## Re: Limit Problems

Originally Posted by Grundsatz
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?

Originally Posted by Grundsatz
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?

Originally Posted by Grundsatz
3. $\underset{x\to \infty }{\mathop{\lim }}\,\frac{\sqrt{x}-1}{x+1}$
This one is easy if you get the first two.

4. ## Re: Limit Problems

Originally Posted by Grundsatz
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*}

5. ## Re: Limit Problems

Originally Posted by Grundsatz
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*}