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:
- $\ds \lim_{x \mathop \to a} \map f x = A$
- $\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in \R: 0 < \size {x - a} < \delta \implies \size {\map f x - A} < \epsilon$
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
- 1961: David V. Widder: Advanced Calculus (2nd ed.) ... (previous) ... (next): $1$ Partial Differentiation: $\S 2$. Functions of One Variable: $2.1$ Limits and Continuity: Example $\text A$