Evaluate each limit. If the limit does not exist due to a vertical asymptote, then add an approach statement stating if $y$ is approaching negative or positive infinity.

1. $\lim\limits _ { x \rightarrow 0 } \large\frac { 1 - \sqrt { x } } { x - 1 }$

   Hint (a):

   Simply evaluate at $x = 0$.

2. $\lim\limits _ { x \rightarrow 1 } \large\frac { 1 - \sqrt { x } } { x - 1 }$

    Strategy 1(b):

   $\lim\limits_{x\rightarrow 1}\frac{1-\sqrt{x}}{x-1}\left ( \frac{1+\sqrt{x}}{1+\sqrt{x}} \right )$

   $\lim\limits_{x\rightarrow 1}\frac{1-x}{(x-1)(1+\sqrt{x})}$

   $\lim\limits_{x\rightarrow 1}\frac{-(x-1)}{(x-1)(1+\sqrt{x})}$

   $\lim\limits_{x\rightarrow 1}\frac{-1}{1+\sqrt{x}}=\underline{ \ \ \ \ \ }$

   You can multiply the top and bottom by the conjugate:

    Strategy 2(b):

   You might have noticed that this is Ana's definition of the derivative:	$\lim\limits_{x\rightarrow a}\frac{f(a)-f(x)}{a-x}$
   Rewrite the given limit as:	$-\lim\limits_{x\rightarrow 1}\frac{\sqrt{x}-1}{x-1}$
   Deconstruct the definition:	$-f(x)=-\sqrt{x}=-x^{1/2}$$-f'(x)=-\frac{1}{2}x^{-1/2}=-\frac{1}{2\sqrt{x}}$
   Write the derivative equation:	$a = 1$
   Evaluate at $x = 1$:	$-f'(1)=\underline{ \ \ \ \ \ \ \ }$

   Answer (b):

   Either way, your answer should be $-\frac{1}{2}$.

3. $\lim\limits _ { h \rightarrow 0 } \large\frac { \sqrt { 25 + h } - 5 } { h }$

   Hint (c):

   This is Hana's definition of the derivative.

   $\lim\limits_{h\rightarrow 0}\frac{f(a+h)-f(a)}{h}$

   Hint (c):

   $f(x)=\sqrt{x}$

   $a = 25$

   Find $f^\prime(x)$ and $f ^\prime(25)$.

4. $\lim\limits _ { h \rightarrow - 2 } \large\frac { ( h + 2 ) - 2 } { 2 }$

   Hint (d):

   Refer to the hint in part (a).

5. Which of the problems above can be interpreted as the definition of the derivative at a point?

   Hint (e):

   Refer to the previous hints.