$\lim_{x\to \infty}e^{-x^2}\int_x^{x+1/x}e^{t^2}dt$

Question

$\lim_{x\to \infty}e^{-x^2}\int_x^{x+1/x}e^{t^2}dt$

This is problem from Spivak calculus 1994, chapter 18, 31.

The book gives following solution: simply apply Lhopitatal rule, but what puzzles me is the way numerator is differentiated.

So they got $e^{(x+1/x)^2}-e^{x^2}\over 2xe^{x^2}$

But I think it should be $(1- 1/x^2) e^{(x+1/x)^2}-e^{x^2}\over 2xe^{x^2}$ . I think they are missing derivative of $x+1/x$. Am I wright or wrong?

P.S Could anyone please help me editing this question?

Answer 1

The function $\;e^{x^2}\;$ is continuous everywhere so it has a primitive function, say $\;F\;$ , and then

$$\int_x^{x+\frac1x}e^{t^2}dt=F\left(x+\frac1x\right)-F(x)\implies$$

$$\lim_{x\to\infty}e^{-x^2}\int_x^{x+\frac1x}e^{t^2}dt=\lim_{x\to\infty}\frac{F\left(x+\frac1x\right)-F(x)}{e^{x^2}}\stackrel{L'H}=\lim_{x\to\infty}\frac{\left(1-\frac1{x^2}\right)e^{\left(x+1/x\right)^2}-e^{x^2}}{2xe^{x^2}}=$$

$$=\lim_{x\to\infty}\frac{\left(\left(1-\frac1{x^2}\right)\left(e^2+e^{1/x^2}\right)\right)-1}{2x}=0\cdot\left((e^2+1)-1\right)\cdot1=0$$

Answer 2

No need to do a complicated derivative.

We can clearly assume $x>0$. Set $g(x)=x+1/x$ for brevity.

By the mean value theorem there exists $c_x\in(x,g(x))$ such that $$ \int_{x}^{g(x)}e^{t^2}\,dt=e^{c_x^2}(g(x)-x)= \frac{e^{c_x^2}}{x} $$ Since $c_x<g(x)$ and the exponential is increasing, the integral is bounded above by $$ \int_{x}^{x+1/x}e^{t^2}\,dt\le\frac{e^{g(x)^2}}{x}=e^2\frac{e^{x^2}e^{1/x^2}}{x} $$ Then $$ 0\le e^{-x^2}\int_{x}^{x+1/x}e^{t^2}\,dt\le e^2\frac{e^{1/x^2}}{x}\le\frac{e^3}{x} $$ By squeezing, the limit is $0$.