Functions (1 Viewer)

khorne · Aug 4, 2011

Prove (e^x-1)/x for x not equal to 0 and f(x) = 1 for x = 0 is continuous. I get i need to prove ls 1 for ( e^x -1)/x but im not exactly sure how without like le'hosp. Theorem

cutemouse · Aug 4, 2011

A function f is cts at a iff

$\lim_{x \to a^-}f(x)=\lim_{x \to a^+} f(x)=\lim_{x \to a}f(x)=f(a)$

Pretty straightforward using this definition...

And btw this isn't MX2. It's 1st year uni stuff.

hscishard · Aug 4, 2011

Don't get your question.
Seems like you can prove it graphically (from what I understood from question)

Shadowdude · Aug 4, 2011

cutemouse said:
A function f is cts at a iff

$\lim_{x \to a^-}f(x)=\lim_{x \to a^+} f(x)=\lim_{x \to a}f(x)=f(a)$

Pretty straightforward using this definition...

And btw this isn't MX2. It's 1st year uni stuff.

^ That.

XTsquared · Aug 4, 2011

cutemouse said:
A function f is cts at a iff

$\lim_{x \to a^-}f(x)=\lim_{x \to a^+} f(x)=\lim_{x \to a}f(x)=f(a)$

Pretty straightforward using this definition...

And btw this isn't MX2. It's 1st year uni stuff.

Afaik that stuff was taught in year 11 (at least using the formula/method posted for determining whether a function was continuous at a given point is year 11).

Shadowdude · Aug 4, 2011

If a function is differentiable, it is continuous.

deterministic · Aug 4, 2011

You mean finding $\lim_{x\rightarrow 0} \frac{e^x-1}{x}=1$ without L'hopital's? Then try using the Taylor expansion of $e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$

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