So, as you know the continuity of a function at a point has some conditions. One of them is that the limit of this function at that point has to exist and be the same as the function's output of that point. Consider the function sqrt(x), which is defined over [0, +inf), if we take the limit of the function at x = 0, the limit would exist from the right side but not the left side. Thus, the full limit wouldn't even exist and yet it's considered continuous at x = 0. How's that?
[Math] Continuity of square root function at the first point
calculuscontinuity
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You are attempting to evaluate the limit of $f(x)$ as $x\to2$ by substituting $x=2$ and giving the answer $f(2)$. In other words, $$\lim_{x\to a}f(x)=f(a)\ .$$ This is true as long as $f(x)$ is continuous at $x=a$. BUT in this case, that is exactly what the question is asking - is $f(x)$ continuous at $x=2$? So you cannot use this approach.
As suggested in other posts, the best way to do this problem is to look at the left and right hand limits separately. You could try thinking about the value of $[x]$ if $x$ is slightly less than, or slightly greater than $2$: $$[x]=\cases{1&if $1\le x<2$\cr 2&if $2\le x<3$\cr}$$ and therefore $$f(x)=x\,[x]=\cases{x&if $1\le x<2$\cr 2x&if $2\le x<3$.\cr}$$ Hence $$\lim_{x\uparrow2}f(x)=\lim_{x\uparrow2}x=2 \quad\hbox{and}\quad \lim_{x\downarrow2}f(x)=\lim_{x\downarrow2}2x=4\ .$$ Since the left and right hand limits are not equal, $\lim_{x\to2}f(x)$ does not exist. Therefore $f(x)$ is not continuous at $x=2$.
Well, that is not the rigurous definition of continuity but it works for most UNDERGRADUATE functions.
Some tricks to check continuity:
- You may know that elementary fucntions ($\sin$, $\cos$, $\exp$, polynomials,...) are continuous everywhere.
- $\log(\cdot)$ is continuous on its domain (whenever what is inside is strictly positive).
- $\sqrt{\cdot}$ is continuous on its domain (wheneever what is inside is positive (not stricly).
- Moreover, any linear combination of continuous functions is also continuous (say, the addition of subtraction, and multiplication by a number). Also multiplication of continuous functions is also continuous.
- For quotient of contuinuous functions, everything works okay EXECEPT for those points that cancel the denominator.
These are just a few tricks; they won’t prove continuity in every case, but for undegraduate students they may be enough.
Best Answer
The limit of a function at a point is defined with respect to its domain. If $D\subseteq\mathbb{R}$, $f:D\to\mathbb{R}$, and $a\in\mathbb{R}$, then we say $L=\lim_{x\to a}f(x)$ if for all $\epsilon>0$ there exists $\delta>0$ such that all $x\color{red}{\in D}$ with $0<|x-a|<\delta$, $|f(x)-L|<\epsilon$. Notice that this definition only involves values of $x$ that actually are in the domain $D$ of the function. So if $f(x)=\sqrt{x}$ with domain $[0,\infty)$, then $\lim_{x\to a}f(x)$ does exist. It's no problem that $f$ is not defined to the left of $0$, since the definition of the limit only cares about inputs that are in the domain of $f$.
(In fact, I would say it's not even correct to say that the limit from the left does not exist. To define the limit from the left, you would use the same definition as above but replace $0<|x-a|<\delta$ by $0<a-x<\delta$. For the case of the square root function, the condition would then hold vacuously, since there are no $x\in D$ such that $0<a-x<\delta$. So actually, it's not that the limit from the left doesn't exist--instead, $L=\lim_{x\to 0^-}\sqrt{x}$ is simultaneously true for all real numbers $L$. In particular, it is true that $0=\lim_{x\to 0^-}\sqrt{x}$, which is what continuity from the left would require. See my answer at The definition of continuity and functions defined at a single point for some related discussion.)