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
please format your question better. i don't know what you are asking
do you know what the definition says?also
If c>0, then lim square root of x = square root of c
x-c
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?
The first one has also been asked and replied to here: http://www.mathhelpforum.com/math-he...ove-limit.html
$\displaystyle | f(x) - l | = | \sqrt{x} - \sqrt{c} | = \left| \frac{(\sqrt{x} - \sqrt{c} ) (\sqrt{x} + \sqrt{c} )}{\sqrt{x} + \sqrt{c} } \right| = \left| \frac{x - c}{\sqrt{x} + \sqrt{c} }\right| = \frac{1}{\sqrt{x} + \sqrt{c}} \, | x - c |$.
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 $\displaystyle 0 < | x - c | < \delta \Rightarrow | \sqrt{x} - \sqrt{c} | < \epsilon$.