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Heyy,

Sorry about putting two questions in single thread, but felt it's easier this way :D

the function in question is

$$ f(x) = \begin{cases} 3x + 4, & \text{if $x \in$ Q} \\ 7x, & \text{if $x \notin$ Q} \end{cases} $$

And the question asks if the following limit exists $$ lim_{x\to 1}f(x) $$

since there can be irrational values of x when x reaches 1 from both sides, i decided that the limit would not exist.

(1) So my first question is whether the limit exists or not

(2) And second is how to plot this function in wolfram alpha so i can visualize the graph? I tried piecewise function plotting but i cannot define the conditions that x is an element of rational/irrational number sets

Thank You!

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  • $\begingroup$ Does there exist a subsequence of real numbers $x_i$ such that $x_i\to 1$ but $f(x_i)\to a$ for some $a\ne 7$? $\endgroup$ – abiessu Mar 12 '15 at 21:02
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Both $3x+4$ and $7x$ are close to $7$ if $x$ is close to $1$. Hence the function is continuous at $1$.

You could try to visualize the graph by drawing two dotted lines representing the linear functions on $\mathbb Q$ and on its complement.

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