As per the definition of limits if lim x → a f ( x ) = L, then ∀ ε &gt; 0 ∃ δ &gt; 0 s . t 0 &lt; | x − a | &lt; δ ⟹ 0 &lt; | f ( x ) − L | &lt; ε

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As per the definition of limits if       lim          x      →      a        f  (  x  )  =  L, then  ∀  ε  &amp;gt;  0     ∃  δ  &amp;gt;  0     s  .  t  0  &amp;lt;  |  x  −  a  |  &amp;lt;  δ       ⟹       0  &amp;lt;  |  f  (  x  )  −  L  |  &amp;lt;  ε

poquetahr · Accepted Answer

A different approach for the sake of curiosity.Let   0  &amp;lt;      |    x  −  a      |    &amp;lt;      δ          ε      . Then we have that:                              |                f        (        x        )        −        L                  |                                    =                  |                          x                      2                          −                  a                      2                                    |                                                                =                  |                (        x        −        a        )        (        x        +        a        )                  |                                                                =                  |                x        −        a                  |                          |                (        x        −        a        )        +        2        a                  |                                                                ≤                  |                x        −        a                  |                (                  |                x        −        a                  |                +        2                  |                a                  |                )                                                        &amp;lt;                  δ                      ε                          (                  δ                      ε                          +        2                  |                a                  |                )        :=        ε            where you can choose the positive root of the corresponding equation on       δ          ε

uplakanimkk · Accepted Answer

Alternative approach:Without loss of generality,   a  &amp;gt;  0. That is, the approach for   a  &amp;lt;  0 is similar, while the approach if   a  =  0 is trivial.If   δ  =  ϵ  , then       |    x  −  a      |    &amp;lt;  δ    ⟹        |    x  −  a      |    &amp;lt;  ϵInstead, take       δ    =    min          (              a        2            ,              ϵ                  3          a                    )      Then,       |    x  −  a      |    &amp;lt;  δ    ⟹        a    2    &amp;lt;  x  &amp;lt;            3      a        2  This implies that       |    x  +  a      |    &amp;lt;            5      a        2    &amp;lt;  3  aSo, you have that       |    x  −  a      |    &amp;lt;  δ and      |    x  +  a      |    &amp;lt;  3  aTherefore      |        x    2    −      a    2        |    =      |    x  −  a      |    ×      |    x  +  a      |    &amp;lt;  δ  ×  3  a  ≤  ϵ

As per the definition of limits if <munder> <mo movablelimits="true" form="prefix">lim <m

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