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Suppose i have two functions $g(x)=\frac{1}{x}$ and $f(x) = \ln x$ and i need to calculate the limit $\lim_{x\to \infty}(g\circ f)(x)$

By composition of limits i can get $\lim_{x\to \infty}f(x)=\infty$, so i will write:

$\lim_{x\to \infty}(g\circ f)(x) = \lim_{x\to \lim_{x\to \infty}f(x)}g\left(f(x)\right) = \lim_{x\to 0}g\left(f(x)\right) = 0$

The result is correct, but i dont know if in general, the following proposition is true or not:

$\lim_{x\to c}g\left(f(x)\right)$ can be calculated as $\lim_{x\to\left[\lim_{x\to c}f(x)\right]}g\left(f(x)\right)$ if $\lim_{x\to c}f(x)$ is $\pm \infty$ or a finite value.

ESCM
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  • You need to remove the second $f$ in ' can be calculated as $\lim_{x\to\left[\lim_{x\to c}f(x)\right]}g\left(f(x)\right)$'. – ir7 Jul 10 '20 at 22:31

2 Answers2

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Not in general. Example: $$ g(x) = \begin{cases} x^2 &, x<1 \\ 0 &, x = 1 \\ 2- x &, x > 1 \end{cases} $$

and $$f(x) = 1 $$ for all $x$.

We take $c=1$. Then $$\lim_{x \rightarrow 1} f(x)=1$$ and

$$ \lim_{x \rightarrow 1} g(f(x)) = 0 \not= 1 = \lim_{y \rightarrow 1} g(y)$$

It is $g$'s (removable) discontinuity at $1$ that makes this difference between the two limits possible.

ir7
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Classical result is when $x_0$ finite or not, is limit point for $f$ domain $X$, $y_0$, finite or not, is limit point for $f(X)$ and $g$ is defined on $f(X)$, then if exists limits, in respect to not punctured neighborhoods, finite or not, $\lim_{x \to x_0}f(x)=y_0$ and $lim_{y \to y_0}g(y)$ then exists $\lim_{x \to x_0}g(f(x))=\lim_{y \to y_0}g(y)$.

If we consider limits against punctured neighborhoods, then additionally we need condition: exists some punctured neighborhood of $x_0$, where $f(x) \ne y_0.$ Then again holds: $$\lim_{x \to x_0}g(f(x))=\lim_{y \to y_0}g(y)$$

zkutch
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