Limit iff Limits from Left and Right
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Theorem
Let $f$ be a real function defined on an open interval $\openint a b$ except possibly at a point $c \in \openint a b$.
Then:
- $\map f x \to l$ as $x \to c$
- $\map f x \to l$ as $x \to c^-$
and
- $\map f x \to l$ as $x \to c^+$
Proof
Necessary Condition
Let $\map f x \to l$ as $x \to c$.
Then from the definition of the limit of a function:
- $\forall \epsilon > 0: \exists \delta > 0: 0 < \size {x - c} < \delta \implies \size {\map f x - l} < \epsilon$
So for any given $\epsilon$, there exists a $\delta$ such that:
- $0 < \size {x - c} < \delta$
implies that:
- $l - \epsilon < \map f x < l + \epsilon$
Now:
\(\ds \) | \(\) | \(\ds 0 < \size {x - c} < \delta\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \) | \(\) | \(\ds - \delta < -\paren {x - c} < 0\) | |||||||||||
\(\ds \lor\) | \(\) | \(\ds 0 < \paren {x - c} < \delta\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \) | \(\) | \(\ds c - \delta < x < c\) | |||||||||||
\(\ds \lor\) | \(\) | \(\ds c < x < c + \delta\) |
That is: $\forall \epsilon > 0: \exists \delta > 0$:
- $(1): \quad c - \delta < x < c \implies \norm {\map f x - l} < \epsilon$
- $(2): \quad c < x < c + \delta \implies \norm {\map f x - l} < \epsilon$
So given that particular value of $\epsilon$, we can find a value of $\delta$ such that the conditions for both:
- $(1): \quad f$ tending to the limit $l$ as $x$ tends to $c$ from the left
and :
- $(2): \quad f$ tending to the limit $l$ as $x$ tends to $c$ from the right.
Thus:
- $\ds \lim_{x \mathop \to c} \map f x = l$
implies that:
- $\ds \lim_{x \mathop \to c^-} \map f x = l$
and:
- $\ds \lim_{x \mathop \to c^+} \map f x = l$
$\Box$
Sufficient Condition
Let $\map f x \to l$ as $x \to c^-$ and $\map f x \to l$ as $x \to c^+$.
This means that:
- $(1): \quad \forall \epsilon > 0: \exists \delta > 0: c - \delta < x < c \implies \size {\map f x - l} < \epsilon$
and :
- $(2): \quad \forall \epsilon > 0: \exists \delta > 0: c < x < c + \delta \implies \size {\map f x - l} < \epsilon$
In the same manner as above, the conditions on $\delta$ give us that:
\(\ds \) | \(\) | \(\ds c - \delta < x < c\) | ||||||||||||
\(\ds \land\) | \(\) | \(\ds c < x < c + \delta\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \) | \(\) | \(\ds 0 < \size {x - c} < \delta\) |
So:
- $\forall \epsilon > 0: \exists \delta > 0: 0 < \size {x - c} < \delta \implies \size {\map f x - l} < \epsilon$
Thus:
- $\ds \lim_{x \mathop \to c^-} \map f x = l$
and:
- $\ds \lim_{x \mathop \to c^+} \map f x = l$
together imply that:
- $\ds \lim_{x \mathop \to c} \map f x = l$
Hence the result.
$\blacksquare$
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 8.4$
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): Appendix: $\S 18.6$: Limits of functions