# How to place a limit that it's inside the integral, outside.

I did this: $$\int_{1}^t x^{-1}dx=\int_{1}^t\lim_{n\rightarrow -1}{x^n}dx =\lim_{n\rightarrow -1}\int_{1}^t{x^n}dx$$ just to have a way to approximate $\ln t$. $$\ln{t}=\lim_{h\rightarrow 0}\frac{x^{h}-1}{h}$$

The second expression may be correct, but I was told I cannot say $\int_a^b \lim_{n\rightarrow -1}x^{n}dx=\lim_{n\rightarrow -1}\int_a^b x^n dx$ without previously prooved some statements. If so, what are those statement and where can I find information about this?

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You can't do that when the exponent is $-1$, can you write $\int x^{-1}=\dfrac{x^{-1+1}}{-1+1}$? –  Inceptio May 15 '13 at 3:58
The result $\ln x=\lim\limits_{h\to0}\frac{x^h-1}{h}$ is true though. Assuming you are allowing yourself the knowledge that $\frac{d}{dh}x^h=x^h\ln x$, this is true by an application of L'Hospital's Rule. –  alex.jordan May 15 '13 at 4:18
I didn't do that @Inceptio . The idea was approching x^-1 to x^-0.9999 so I could integrate. In fact, you could get a nice approximation of $\ln x$ with $\frac{x^{0.00001}-1}{0.00001}$. –  Alan May 15 '13 at 4:21

In general, you cannot interchange two limits. In this case, recall that the integration is a actually a limit, for instance, the limit of a Riemann sum. Hence, you need to be careful, in general: $$\int_a^b \lim_{n \to \infty} f_n(x) dx \neq \lim_{n \to \infty} \int_a^b f_n(x) dx$$