  ### Home > APCALC > Chapter 3 > Lesson 3.3.2 > Problem3-110

3-110.

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

Simply evaluate at $x = 0$.

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

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

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

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

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

You can multiply the top and bottom by the conjugate:

 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 $x = 1$: $-f'(1)=\underline{ \ \ \ \ \ \ \ }$

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

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

This is Hana's definition of the derivative.

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

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

$a = 25$

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

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

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?

Refer to the previous hints.