1. ## Finding limits

2. Can you show your hours of working?

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

3. Originally Posted by letzdiscuss
I'm numbering you're problems:

1. 2.
3. 4.
5. 6.
7.

1.) Note that $\displaystyle \frac{x}{|x|}=sign(x)=\left\{\begin{array}{lr}1&:x >0\\0&:x=0\\-1&:x<0\end{array}\right\}$

Spoiler:
So $\displaystyle \frac{x^2}{|x|}=sign(x)\cdot x$ and $\displaystyle \lim_{x\to0}(sign(x)\cdot x)=0$

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2.) This function is continuous (from the right) at $\displaystyle x=1$.

Spoiler:
Just plug it in.

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3.) Just plug it in.

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4.) Start by factoring the numerator and see where that gets you.
Spoiler:
$\displaystyle x^3-8=(x-2)(x^2+2x+4)$. Cancel out the $\displaystyle (x-2)$ terms, and plug in $\displaystyle x=2$.

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5.) Same strategy as the last one.

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6.) As $\displaystyle x$ gets small, $\displaystyle \frac{1}{|x|}$ gets really big, so $\displaystyle \lim_{x\to0}\frac{1}{|x|}=\infty$.

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7.) Same method as (4) and (5).

4. Originally Posted by redsoxfan325
I'm numbering you're problems:

1. 2.

3. 4.

5. 6.

7.

1.) Note that $\displaystyle \frac{x}{|x|}=sign(x)=\left\{\begin{array}{lr}1&:x >0\\0&:x=0\\-1&:x<0\end{array}\right\}$

Spoiler:
So $\displaystyle \frac{x^2}{|x|}=sign(x)\cdot x$ and $\displaystyle \lim_{x\to0}(sign(x)\cdot x)=0$

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2.) This function is continuous (from the right) at $\displaystyle x=1$.

Spoiler:
Just plug it in.

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3.) Just plug it in.

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4.) Start by factoring the numerator and see where that gets you.
Spoiler:
$\displaystyle x^3-8=(x-2)(x^2+2x+4)$. Cancel out the $\displaystyle (x-2)$ terms, and plug in $\displaystyle x=2$.

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5.) Same strategy as the last one.

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6.) As $\displaystyle x$ gets small, $\displaystyle \frac{1}{|x|}$ gets really big, so $\displaystyle \lim_{x\to0}\frac{1}{|x|}=\infty$.

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7.) Same method as (4) and (5).

Thanks !! Got it !!