Answer: D) 1/e but you probably know already
Step-by-step explanation:
The Fundamental Theorem of Calculus part 1:
[tex]\text{Suppose g(x) is continuous on [a,b]} \\\text{Then for all x in [a,b]: }\frac{d}{dx}\int\limits^x_ag(t)dt=g(x)\text{*}\\\text{In this case }g(x)=e^{-x^2}\text{ which is continuous over }[-\infty,\infty]\text{**}\\\text{So it's continuous at x=1 in particular}. \text{ Set a=1 and }b=\infty\\\text{We see that }\frac{d}{dx} f(x)=\frac{d}{dx}\int\limits^x_a {e^{-t^2}} \, dt=e^{-x^2}=g(x)\text{ for all x in }[1,\infty][/tex]
A similar argument where a= -infinity and b=1 gives you d/dx f(x) = g(x) for all x in [-infinity, 1]. Therefore it holds for all x in [-infinity, infinity]
Also, d/dx (x+1) = 1
Also also, f(1)=0 because the bounds become equal
Thus the limit is an indeterminate form of 0/0; this suggests using L'Hopital's Rule
[tex]\lim_{x \to 1} \frac{f(x)}{x+1}= \lim_{x \to 1} \frac{\frac{d}{dx} f(x)}{\frac{d}{dx} (x+1)}=\lim_{x \to 1} e^{-x^2}[/tex]
By definition of continuity and the fact that the above g(x) converges at 1, we have
[tex]= e^{-1^2}=\frac{1}e[/tex]
*there's a proof on proofwiki.org that's too long to fit here, but you probably don't need it
**it's 3am, maybe I'll prove it in the morning if I have time lol. Anyway if want to do it yourself, you can use the limit definition of congruence, drag the limit symbol to the exponent -x^2, etc. Or just apply the theorem without checking for continuity. Tell me if you need more detail (or less rambling)
