$$\lim_{n\rightarrow \infty }\frac{1}{n}\int_{1}^{n}\frac{x-1}{x+1}dx$$

My approach is not correct, I think.

I took $f(x)=(x-1)/(x+1)$ which is continuous so there is a $F(x)$ a primitive of f(x) such that $F'(x)=f(x)$

So I used L'Hospital but when I derived, I derived for n which is number, not for x and I think it's not correct.

I obtained $\lim_{n \to \infty }\frac{F(n)-F(1)}{n}=\lim_{n \to \infty }f(n)=1$ which is the correct answer but I'm not sure if this is the right method.There's another way to solve this?

  • 2
    $\begingroup$ Using L'Hopital's Rule is the natural approach and your answer is correct. $\endgroup$ – Kavi Rama Murthy Apr 21 '19 at 12:02

Yes, your approach is correct. You applied L'Hôpital's rule to $$ \lim_{x\to \infty } \frac{F(x) - F(1)}{x} = \lim_{x\to \infty } \frac{F'(x)}{1} = 1 $$ which is valid because the denominator $g(x) = x $ on the left-hand side diverges to $ \infty$, satisfies $g'(x) \ne 0$, and the limit on the right-hand side exists.

Then $ \frac{F(x_n) - F(1)}{x_n} \to 1$ for all sequences $(x_n)$ with $x_n \to \infty$, and in particular $$ \lim_{n\to \infty } \frac{F(n) - F(1)}{n} = 1 \, . $$

As mentioned in L'Hôpital's rule: General proof (case 2) this is actually a consequence of the Stolz–Cesàro theorem: In your case, $$ a_n = \int_1^n f(x) \, dx \, , \, b_n = n $$ satisfy the hypotheses of the Stolz–Cesàro theorem with $$ \begin{align} \frac{a_{n+1} - a_n}{b_{n+1} - b_n} &= \int_n^{n+1} f(x) \, dx \\ &= f(\xi_n) \text{ for some } \xi_n \in (n, n+1) \\ &\to 1 \text{ for }n \to \infty \, , \end{align} $$ using the mean value theorems for integrals. It follows that $$ \frac 1n \int_1^n f(x) \, dx = \frac{a_n}{b_n} \to 1 $$ for $n \to \infty$.

  • $\begingroup$ L'Hospital's Rule applies to the form $\text{anything} /\infty$ so verifying that $F(x) \to\infty $ is not needed. $\endgroup$ – Paramanand Singh Apr 21 '19 at 15:19
  • $\begingroup$ @ParamanandSingh: You are right, thanks. I have removed that part. $\endgroup$ – Martin R Apr 21 '19 at 15:31
  • $\begingroup$ Thank you for your answer! $\endgroup$ – DaniVaja Apr 21 '19 at 17:22

Usethat $$\frac{x-1}{x+1}=1-\frac{2}{x+1}$$

  • $\begingroup$ Thank you for your help :) $\endgroup$ – DaniVaja Apr 21 '19 at 12:20

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