Thread: Limits

Evaluate.

1. $\displaystyle \lim_{x \to \--2} \frac{x + 4}{x + 3} = 2$

2. $\displaystyle \lim_{x \to \--2} \frac{x^2 + x - 2}{x^2 + 5x + 6} = -3$

3. $\displaystyle \lim_{x \to \-2}$ $\displaystyle \frac {\sqrt {(x-1)} - 1}{(x - 2)} = 0.5$

4. $\displaystyle \lim_{x \to \-1} \frac{x - 1}{x^2 - 1} = 0.5$

Are my answers correct?

Originally Posted by Macleef

Evaluate.

1. $\displaystyle \lim_{x \to \--2} \frac{x + 4}{x + 3} = 2$

2. $\displaystyle \lim_{x \to \--2} \frac{x^2 + x - 2}{x^2 + 5x + 6} = -3$

3. $\displaystyle \lim_{x \to \-2}$ $\displaystyle \frac {\sqrt {(x-1)} - 1}{(x - 2)} = 0.5$

4. $\displaystyle \lim_{x \to \-1} \frac{x - 1}{x^2 - 1} = 0.5$

Are my answers correct?

1.yes
2.yes
3.yes
4.yes

1. $\displaystyle \lim_{x\to -2}\frac{x + 4}{x + 3}$

$\displaystyle = \frac{-2 + 4}{-2 + 3} = \frac{2}{1} = 2$

2. $\displaystyle \lim_{x \to \--2} \frac{x^2 + x - 2}{x^2 + 5x + 6}$

$\displaystyle = \lim_{x \to \--2} \frac{2x + 1}{2x + 5}$

$\displaystyle = \frac{-4 + 1}{-4 + 5} = \frac{-3}{1} = -3$

3. $\displaystyle \lim_{x\to 2}\frac {\sqrt {(x-1)} - 1}{(x - 2)}$

$\displaystyle = \lim_{x \to 2}\frac{1}{2}(x-1)^{-\frac{1}{2}}$

$\displaystyle = \lim_{x \to 2}\frac{1}{2\sqrt{(x -1)}}$

$\displaystyle = \frac{1}{2\sqrt{(2 - 1)}}$

$\displaystyle = \frac{1}{2}$

4. $\displaystyle \lim_{x \to 1} \frac{x - 1}{x^2 - 1}$

$\displaystyle = \lim_{x \to 1} \frac{1}{2x}$

$\displaystyle = \frac{1}{2}$

So, yes to all.