### 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 $x = 0$.

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

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

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

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

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

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 $x = 1$:

$\lim \limits_{x\rightarrow a}\frac{f(a)-f(x)}{a-x}$

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

$-f(x)=-\sqrt{x}=-x^{1/2}$

$-f'(x)=-\frac{1}{2}x^{-1/2}=-\frac{1}{2\sqrt{x}}$

$a = 1$

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

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

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

This is Hana's definition of the derivative.

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

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

$a = 25$

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

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

Refer to the hint in part (a).

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

Refer to the previous hints.