A Limit Problem

Calculus Level 4

$\large{ \lim_{x \to 0} \dfrac{1}{x} \int_x^{2x} e^{-t^2} \, dt }$

Determine if the limit above exists or not. If it does exist, find its value.

Exists and equals 0 Exists and equals 1 It does not exist because it diverges to infinity It does not exist because it doesn't diverge to any value

1 solution

Brian Charlesworth
May 10, 2017

Let $\displaystyle F(x) = \large \int_{x}^{2x} e^{-t^{2}} dt$ . Then as $\large e^{-x^{2}}$ is continuous (and differentiable) over $\mathbb{R}$ we see that $\displaystyle \lim_{x \to 0} F(x) = 0$ , and so the given limit $\displaystyle \lim_{x \to 0} \dfrac{F(x)}{x}$ is of the indeterminate form $0/0$ . As such, we can apply L'Hopital's rule, along with the Fundamental Theorem of Calculus, to find that

$\large \displaystyle \lim_{x \to 0} \dfrac{F(x)}{x} = \lim_{x \to 0} \dfrac{F'(x)}{1} = \lim_{x \to 0} (2e^{-(2x)^{2}} - e^{-x^{2}}) = 2 - 1 = \boxed{1}$ .

Well, I did exactly what you've done but then I thought of doing it in another manner. I assumed the same F(x) as you did. Then I thought, "The value of this integral is always positive, whether x tends to 0+ or 0- because it simply denotes the area underneath the graph of F(x). Now, when x tends to 0-, we're dividing a positive quantity by a negative quantity, which would yield a negative number (if the value of the limit is finite) and when x tends to 0+, we're dividing a positive quantity by a positive quantity, which would yield a positive number (again, if the limit is finite). I'm not able to understand where I went wrong. Could you please help?

Vishesh arora - 4 years ago

When $x \to 0^{-}$ , $F(x)$ is actually a negative quantity, since $2x \lt x$ in this case. That is, $F(x)$ is only a positive when the lower bound is less than the upper bound, which is not the case when $x \lt 0$ . So whether we approach $0$ from the left or the right, $\dfrac{F(x)}{x}$ will always be a positive quantity, since when $x \gt 0$ we are dividing a positive by a positive, and when $x \lt 0$ we are dividing a negative by a negative.

Brian Charlesworth - 4 years ago

A Limit Problem

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