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

3-110.

Evaluate each limit. If the limit does not exist, say so but also state if $y$ is approaching positive or negative infinity.

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

Simply evaluate at $x = 0$.

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

$\lim \limits_{x\rightarrow 1}\frac{1-\sqrt{x}}{x-1}\left ( \frac{1+\sqrt{x}}{1+\sqrt{x}} \right )$
$\lim \limits_{x\rightarrow 1}\frac{1-x}{(x-1)(1+\sqrt{x})}$
$\lim \limits_{x\rightarrow 1}\frac{-(x-1)}{(x-1)(1+\sqrt{x})}$
$\lim \limits_{x\rightarrow 1}\frac{-1}{1+\sqrt{x}}=\underline{ \ \ \ \ \ }$
You can multiply the top and bottom by the conjugate:
You might have noticed that this is Ana's definition of the derivative:

Rewrite the given limit as:
Deconstruct the definition:
Write the derivative equation:
Evaluate at $x = 1$
:$\lim \limits_{x\rightarrow a}\frac{f(a)-f(x)}{a-x}$
$-\lim \limits_{x\rightarrow 1}\frac{\sqrt{x}-1}{x-1}$
$-f(x)=-\sqrt{x}=-x^{1/2}$$-f'(x)=-\frac{1}{2}x^{-1/2}=-\frac{1}{2\sqrt{x}}$
$a = 1$
$-f'(1)=\underline{ \ \ \ \ \ \ \ }$

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

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

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 $f(x)$ and $f '\left(25\right)$.

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

Refer to the hint in part (a).

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

Refer to the previous hints.