Limit of Real Function/Examples/Root of x at 1

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Example of Limit of Real Function

$\ds \lim_{x \mathop \to 1} \sqrt x = 1$


Proof

By definition of the limit of a real function:

$\ds \lim_{x \mathop \to a} \map f x = A$

if and only if:

$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in \R: 0 < \size {x - a} < \delta \implies \size {\map f x - A} < \epsilon$


In this instance, we have:

$\map f x = \sqrt x$
$A = 1$

So:

\(\ds \size {\map f x - A}\) \(=\) \(\ds \size {\sqrt x - 1}\)
\(\ds \) \(=\) \(\ds \dfrac {\size {x - 1} } {\sqrt x + 1}\) multiplying top and bottom by $\sqrt x + 1$

for $0 < x < 2$.

Let $\delta = \epsilon$.

Let $0 < \size {x - 1} < \delta = \epsilon$.

Then we obtain:

\(\ds \size {\sqrt x - 1}\) \(<\) \(\ds \dfrac \epsilon {\sqrt x + 1}\)
\(\ds \) \(<\) \(\ds \epsilon\) multiplying top and bottom by $\sqrt x + 1$

$\blacksquare$


Sources