Is ${\lim\limits_{x \to 0}} \frac{f(x^2+1)}{x^2+2} = {\lim\limits_{t \to 1}} \frac{f(t)}{2}$ always true?

Question

Is ${\lim\limits_{x \to 0}} \frac{f(x^2+1)}{x^2+2} = {\lim\limits_{t \to 1}} \frac{f(t)}{2}$ always true?

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My math teachers were arguing today if the statement is always true. One teacher gave a counterexample of where $f(x) = \sqrt{x-1}$. Eventually, they said that it came down to different textbook definitions of continuity and limits. Is there any other reasoning to prove this to be true or false?

Answer 1

Your limit approach is off.

$$\lim_{x \to 0} \frac{f(x^2+1)}{x^2+2} \neq \lim_{t \to 1} \frac{f(t)}{2} $$

Rather

$$\lim_{x \to 0} \frac{f(x^2+1)}{x^2+2} = \lim_{t \to 1^+} \frac{f(t)}{2} $$

Since as $x\to 0$, $(x^2+1) \to 1$, but also $x^2+1 > 1$ for all $x\neq 0$.

It doesn't make sense to write

$$ \lim_{t \to 1^-} \frac{f(t)}{2} $$

in this case.

Answer 2

Consider for example a piecewise function $f(x) = 0$ for $x<1$ and $f(x) = 1$ for $x \ge 1$. Then the left hand side is defined and equal to $1/3$, but the right hand side is not defined.

If right hand limit exists then indeed they are equal though.

Answer 3

Yes, it does come down to the definition of continuity and limits, but most imporantly it comes down to the definition of a function. Some functions are not defined everywhere. A good example is $\ f(x) = \sqrt{x}\ $ which is well defined as a real valued function only if $\ x\ $ is a non-negative real number. The usual definition of limits only require using the values in the domain of a function. In this particular case, the domain is $\ [0,\infty).\ $ Thus, by usual definition, $\ f()\ $ is continuous at $\ x=0\ $ because $\ \lim_{x\to 0} f(x) = \lim_{x\to 0+} f(x) = 0.\ $

For the example in the question the left limit only uses values of $\ f(x)\ $ for $\ x\ge 1\ $ and the function may have arbitrary values for $\ x<1.\ $ The right limit appears to be a two-sided limit at $\ x=1.\ $ If the domain of the function $\ f()\ $ was $\ [1,\infty)\ $ then the two limits would be the same, but without knowing the domain of the function and its values for $\ x<1\ $ it is not valid to equate the two limits.

Answer 4

$${\lim_{x \to 0}} \frac{f(x^2+1)}{x^2+2} \overset{?}{=} {\lim\limits_{t \to 1}} \frac{f(t)}{2}$$

• If $f$ is continuous at $b$ and $\displaystyle \lim_{x \to a}g(x)=b, \quad$ then $\quad\lim_{x \to a}f(g(x))=f(\lim_{x \to a}g(x)) = f(b)$.

Let $F(x) = \dfrac{f(x)}{x+1}$. Then $F(x)$ is continuous at $x=0$ if and only if $f(x)$ is continuous at $x=1$.

So, suppose that $\mathbf{\text{$\mathbf{f(x)}$ is continuous at $\mathbf{x=1}$}}$.

Let $G(x)=x^2+1$. Then $${\lim_{x \to 0}} \frac{f(x^2+1)}{x^2+2} = \displaystyle \lim_{x \to 0}F(G(x)) =F(\lim_{x \to 0}G(x)) = \dfrac{f(1)}{2} = \lim_{t \to 1}\frac{f(t)}{2}$$