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Show that the function

$ f(x) = \left\{

\begin{array}{ll}

x^4 \sin (1/x) & \mbox{if $ x \neq 0 $}\\

0 & \mbox{if $ x = 0 $}

\end{array} \right.$

is continuous on $ (-\infty, \infty) $.

$f(x)=x^{4} \sin (1 / x)$ is continuous on $(-\infty, 0) \cup(0, \infty)$ since it is the product of a polynomial and a composite of a

trigonometric function and a rational function. Now since $-1 \leq \sin (1 / x) \leq 1,$ we have $-x^{4} \leq x^{4} \sin (1 / x) \leq x^{4}$. Because $\lim _{x \rightarrow 0}\left(-x^{4}\right)=0$ and $\lim _{x \rightarrow 0} x^{4}=0,$ the Squeeze Theorem gives us $\lim _{x \rightarrow 0}\left(x^{4} \sin (1 / x)\right)=0,$ which equals $f(0) .$ Thus, $f$ is continuous at 0 and, hence, on $(-\infty, \infty)$.

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This is problem number seventy one of this tour calculus eighth edition, Section two point five show that the function f of X equals except forthe time, sign of the quantity one over X X is not equal to zero and zero if X is equal to zero, is continuous on the domain of all real numbers and our we're going to use our definition of continuity to confirm our continuity. To make sure that we have confirmed continuity, Dad, for a function f and function F is only considered continuous if and only if the limit is X approaches. EVA function is equal to the function evaluated at eight. So if we take a look at the function, it is definitely continuous on all rials as is on this first function is a combination are two continuous functions polynomial and a trigonometry function the sign of or the quantity one of Rex is on ly discontinuous, where X is equal to zero. However, that's not included in the domain. So at the moment it is definitely continuous on all riel numbers. The issue is at X equals zero. We have to make sure that this corresponds this f of X equals zero corresponds to the limit. As dis approaches zero s O to confirm, we want to make sure that the limit is experts is zero out of the function except forthe time, Sign of the quantity. One Rex equals F zero, which is here on this case. Andi, Until we confirm that we cannot say that this function is continuous on all real numbers. Once we confirmed this, then we can say it. It is continuous from negative infinity to infinity. So what we do is we approach this using the squeeze Terram. We know that the sine function its value. Its range is from negative one to one. And if we want to buy exit the fourth to each turn, this gives us the function in question that we want. Now we take the entire and equality said here and take the limit as X approaches zero. And if you notice the limited express zero of negative X of the fourth will be zero. The limit as X approaches zero of X to the fourth will be zero. And so this limit of this function has experts. Zero must be between zero and zero, and the only Whether that is true is if dysfunction is lim is also equal to zero. So we have taken care of this limit and shown that it is equal to zero and since it is equal to zero, it is equal to zero confirming and that has X approaches. Zero. This function is continuous at X equals zero, and thus the function is continuous from negative infinity to infinity.