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# Prove that $f$ is continuous at $a$ if and only if $$\lim_{h \to 0}f(a + h) = f(a)$$

## $(\Rightarrow)$ If $f$ is continuous at $a,$ then by Theorem 8 with $g(h)=a+h,$ we have$\lim _{h \rightarrow 0} f(a+h)=f\left(\lim _{h \rightarrow 0}(a+h)\right)=f(a)$$(\Leftarrow)$ Let $\epsilon>0 .$ since $\lim _{h \rightarrow 0} f(a+h)=f(a),$ there exists $\delta>0$ such that $0 < |h| < \delta \Rightarrow$$|f(a+h)-f(a)|<\varepsilon . \text { So if } 0< |x-a|< \delta, \text { then }|f(x)-f(a)|=|f(a+(x-a))-f(a)|< c$Thus, $\lim _{x \rightarrow a} f(x)=f(a)$ and $s o f$ is continuous at $a$

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this phone number sixty three of the Stuart Calculus eighth edition section two point five prove that if his continues at a if, and only if the limit is six. H approaches zero of the function April's Each is equal to that of a and if we were calling a function F is on ly. Continuous is continuous if and only if it's limit. As XO. Purchasing of the function is equal to have a mai. So we're going to use this. See if we can associate Thies, too. We're gonna let h equal explains saying. Ah, and if that's true, we can re writes this part and the limit as the limit of the function. April's a jewel. A procedure is the same. A saint ex or let me replace it. And that is a plus age which is experiencing. Cols Asperger's and then a We can ah keep the same over here. No, since ages explain to say we're going to show it as explains a approaching zero. And if we do some more some modifications here a man is a zero, and then here we can add it to both sides of this Aargh. So this would be the same as limited as expert as a or the function f of X because after today so essentially what we have is that at the beginning this is another way of saying the definition of continuity. It is written, written using ancient set of X. Using this conversion, H is equal to x one and saying so we were able to show that this is equivalent to the definition of continuity and thus it is continuous, Ari, for this reason.

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