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$$f(x) = \left\{\begin{array}{ll} x^2 + 2 & : x < 1\\ 2 & : x = 1\\ (x - 1)^2 + 4 & : x> 1 \end{array} \right.$$

Why is the following limit equal to 3? $$\lim\limits_{x \to 0} \ f(1 - x^2)$$ My first thought was that it was nonexistent.

4 Answers 4

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Note that $x^2 > 0$ for $x \neq 0$, so if set $1-x^2=t$, we have $t < 1$.

So we have that $$\lim_{x \to 0} f(1-x^2)=\lim_{t \to 1^{-}}f(t)=\lim_{t \to 1^{-}}(t^2+2)=3$$ As $f(t)=t^2+2$ for $t<1$.

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    Ah! Tricky $x^2$, caught me by surprise! – 2017-02-19
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    So had the composition been 1-x instead of 1-x^2, the limit would be from both sides of 1 and it would then be nonexistent? – 2017-02-19
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    @user1 Yes. It would be so then. – 2017-02-19
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    Can we not just plug in as I have done below? – 2017-02-19
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    @NickPavini: It's the same thing, but S.C.B's answer makes it more explicit why we are able to say that $f(t)=t^2+2$. – 2017-02-19
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$$\lim_{x \to 0}f(1-x^2)=\lim_{x\to0}(1-x^2)^2+2=(1-0)^2+2=3$$ Because $x\to0$, we can use the $x<1$ function and just plug in.

EDIT: The above statement implies, but we must explicitly say that for all $x\not=0$, $x^2>0$ therefore, $1-x^2<1$

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    Why is $$\lim_{x \to 0}f(1-x^2)=\lim_{x\to0}1-x^2+2$$? This is not true. – 2017-02-19
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    @S.C.B. because by the definition of a limit, $x$ is suficiently close to $a$ and $a$ is $0$, approaching $0$ from either direction is $<1$ – 2017-02-19
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    What does that have to do with anything? I meant it should be $$(1-x^2)^2+2$$ – 2017-02-19
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    You are absoulutely right! haha Thank You! @S.C.B. – 2017-02-19
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    @S.C.B.My method still holds with that edit tho no? – 2017-02-19
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    Yes, but it is essentially what I did in my answer, except without the substitution. – 2017-02-19
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    @S.C.B. Ok I am just checking, because sometimes your answers are really smart and hard to understand! – 2017-02-19
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    This answer is correct as far as it goes (in the sense that all the equivalences indeed hold), but it's missing an explicit justification of *why* we can replace $f(1-x^2)$ with $(1-x^2)^2+2$ and ignore the other parts of the definition of $f$ when computing the limit as $x \to 0$. Still, have a +1. – 2017-02-19
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    @IlmariKaronen ok I see what you are saying. That makes a lot of sense!:) – 2017-02-19
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The limit of a function at a given point is determined by the values it takes near, but not at, that point. For example, to determine the limit of $x \mapsto f(1 - x^2)$ at zero, we need to examine the values of $f(1 - x^2)$ at values of $x$ that are close to but not equal to zero.

Assuming that $x$ is meant to be restricted to real numbers (which the piecewise definition of $f$ implicitly suggests), $x^2 > 0$ (and therefore $1 - x^2 < 1$) whenever $x \ne 0$. Thus: $$f(1 - x^2) = (1 - x^2)^2 + 2 \quad \forall x \ne 0.$$

The limit of $(1 - x^2)^2 + 2$ as $x \to 0$ equals $3$.

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The first essential step in evaluating the value or form of a composite function in the form of $f(g(x))$, is to determine on which intervals of the domain of $f(x)$ does $g(x)$ apply(i.e where the domain of $f(x)$ coincides with the range of $g(x)$). In this case with a piece wise function we need to evaluate the range of $g(x)$, as this range is what will determine the interval of the domain of $f(x)$ to consider. Notice that $g(x)=1-x^2$ and $1-x^2<1$ for all values of x except $x=0$. Thus, the domain of f(x) in question is x<1 which is where $f(x)=x^2 + 2$. Consequently $$f(g(x))=(1-x^2)^2+2$$ and $$\lim_{x\to 0}\,f(g(x))=\lim_{x\to 0}\,(1-x^2)^2+2=(1)^2+2=3$$