# Thread: use LIMIT definition

1. ## use LIMIT definition

i need help

with the use of the limit definition how can i prove this

lim 1 = 1 c is not zero
x-c x c

also
If c>0, then lim square root of x = square root of c
x-c

2. Originally Posted by loop
i need help

with the use of the limit definition how can i prove this

lim 1 = 1 c is not zero
x-c x c
please format your question better. i don't know what you are asking

also
If c>0, then lim square root of x = square root of c
x-c
do you know what the definition says?

note that $\displaystyle |\sqrt x - \sqrt c| = \left| (\sqrt x - \sqrt c) \cdot \frac {\sqrt x + \sqrt c}{\sqrt x + \sqrt c} \right| = \frac {|x - c|}{\sqrt x + \sqrt c} \le \frac {|x - c|}{\sqrt c}$

for $\displaystyle c \ne 0$

now can you fill in the blanks?

3. Originally Posted by loop
i need help

with the use of the limit definition how can i prove this

lim 1 = 1 c is not zero Mr F says: I assume you mean $\displaystyle {\color{red}\lim_{x \rightarrow c} \frac{1}{x} = \frac{1}{c}, ~ c \neq 0}$
x-c x c

also
If c>0, then lim square root of x = square root of c Mr F says: I assume you mean $\displaystyle {\color{red}\lim_{x \rightarrow c} \sqrt{x} = \sqrt{c}, ~ c > 0}$
x-c
..

4. yes that is what am trying to say.

5. Originally Posted by loop
sorry about the post. and yes that is what am trying to say.
for the first, we want $\displaystyle \left| \frac 1x - \frac 1c \right| = \left| \frac {c - x}{xc} \right| < \epsilon$ for any $\displaystyle \epsilon > 0$, provided $\displaystyle |x - c| < \delta$ for some $\displaystyle \delta > 0$.

can you fill in the blanks?

6. Originally Posted by loop
i need help

with the use of the limit definition how can i prove this

lim 1 = 1 c is not zero
x-c x c

also
If c>0, then lim square root of x = square root of c
x-c
The first one has also been asked and replied to here: http://www.mathhelpforum.com/math-he...ove-limit.html

7. Originally Posted by loop
[snip]
If c>0, then lim square root of x = square root of c
x-c
$\displaystyle | f(x) - l | = | \sqrt{x} - \sqrt{c} | = \left| \frac{(\sqrt{x} - \sqrt{c} ) (\sqrt{x} + \sqrt{c} )}{\sqrt{x} + \sqrt{c} } \right| = \left| \frac{x - c}{\sqrt{x} + \sqrt{c} }\right| = \frac{1}{\sqrt{x} + \sqrt{c}} \, | x - c |$.

Only values of $\displaystyle x$ in the neighbourhood of $\displaystyle c$ are of interest so impose the restriction $\displaystyle 0 < x < 2c \Rightarrow 0 < | x - c| < c$, say.

Under this restriction, $\displaystyle \frac{1}{\sqrt{x} + \sqrt{c}} < \frac{1}{\sqrt{c}}$. Therefore:

$\displaystyle | f(x) - l | = \frac{1}{\sqrt{x} + \sqrt{c}} \, | x - c | < \frac{1}{\sqrt{c}} | x - c |$.

Note that $\displaystyle | f(x) - l | < \frac{1}{\sqrt{c}} | x - c | < \epsilon \Rightarrow | x - c| < \sqrt{c} \, \epsilon$.

So for all $\displaystyle \epsilon > 0$ there exists $\displaystyle \delta (\epsilon) = \min\{ c, \, \sqrt{c} \, \epsilon\}$ such that $\displaystyle 0 < | x - c | < \delta \Rightarrow | \sqrt{x} - \sqrt{c} | < \epsilon$.