Interchanging a limit and a parametric improper integral

Question

Suppose I have the limit: $$\lim _{t\to 0^+}\int _0^{\sqrt{\sqrt{t}+4}}\sqrt{1+\frac{t}{4\sqrt{x}}}\:dx$$ How can I prove that the limit is $2$? It is easy to prove that for all $t > 0$ the improper inegral converges by using the limit comparison test (the problem is at x = 0). I also see that the upper bound approaches $2$. If I try to write the improper integral as a limit of a proper integral I get: $$\lim _{t\to 0^+}\:\left(\lim _{a\to 0^+}\int _a^{\sqrt{\sqrt{t}+4}}\sqrt{1+\frac{t}{4\sqrt{x}}}\:dx\right)$$ However, I don't know how to deal with an iterated limit. When can I switch the order? Is it even the proper way to tackle the limit? Note that I haven't learned measure theory so I would prefer a solution which doesn't require it.

Answer 1

To prove the limit is $2$ we must show that for any $\epsilon >0$ there exists $\delta > 0$ such that if $0 < t < \delta$, then

$$\tag{*}\left|\int _0^{\sqrt{\sqrt{t}+4}}\sqrt{1+\frac{t}{4\sqrt{x}}}\,dx -2 \right| < \epsilon$$

Note that

$$\left|\int _0^{\sqrt{\sqrt{t}+4}}\sqrt{1+\frac{t}{4\sqrt{x}}}\,dx -2 \right| \leqslant \left|\int _2^{\sqrt{\sqrt{t}+4}}\sqrt{1+\frac{t}{4\sqrt{x}}}\,dx + \int _0^{2}\sqrt{1+\frac{t}{4\sqrt{x}}}\,dx -2 \right|\\ \leqslant \underbrace{\left|\int _2^{\sqrt{\sqrt{t}+4}}\sqrt{1+\frac{t}{4\sqrt{x}}}\,dx\right|}_{A} + \underbrace{\left|\int _0^{2}\sqrt{1+\frac{t}{4\sqrt{x}}}\,dx -2 \right|}_{B}$$

For the term $A$, with $x \geqslant 2$ we have

$$A=\left|\int _2^{\sqrt{\sqrt{t}+4}}\sqrt{1+\frac{t}{4\sqrt{x}}}\,dx\right|\leqslant \sqrt{1+\frac{t}{4\sqrt{2}}}\left(\sqrt{\sqrt{t}+4}-2\right)$$

Since the RHS depends continuously on $t$ there exists $\delta_1>0 $ such that if $0< t < \delta_1$, then $A < \epsilon/2$.

Term $B$ is easily handled by showing that the improper integral converges uniformly for $t$ in any bounded interval $[0,C]$. Since $\left|\sqrt{1+\frac{t}{4\sqrt{x}}}\right| \leqslant \sqrt{1+\frac{C}{4\sqrt{x}}}$ it follows by the Weierstrass M-test that the improper integral appearing in $B$ is uniformly convergent and, thus,

$$\lim_{t \to 0+}\int _0^{2}\sqrt{1+\frac{t}{4\sqrt{x}}}\,dx = \int_0^2\lim_{t \to 0+}\sqrt{1+\frac{t}{4\sqrt{x}}}\,dx = 2$$

Hence, there exists $\delta_2 > 0$ such that when $0 < t < \delta_2$ we have

$$B = \left|\int _0^{2}\sqrt{1+\frac{t}{4\sqrt{x}}}\,dx -2 \right| < \frac{\epsilon}{2}$$

Therefore, we conclude the limit is $2$ since if $0 < t < \delta = \min(\delta_1,\delta_2)$, then

$$\left|\int _0^{\sqrt{\sqrt{t}+4}}\sqrt{1+\frac{t}{4\sqrt{x}}}\,dx -2 \right| \leqslant A + B < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon,$$

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Here we have used a well-known theorem:

If $f:(a,b] \times S \to \mathbb{R}$ is continuous and the improper integral $$F(t) =\int_a^b f(x,t)\,dx = \lim_{c \to a+}\int_c^b f(x,t) \, dx$$ is uniformly convergent for $t \in S$, then $F:S \to \mathbb{R}$ is continuous on $S$.

To prove this we consider the sequence $(F_n)$ given by

$$F_n(t) = \int_{a + 1/n}^b f(x,t) \, dt$$

Then $F_n(t) \to F(t)$ and the covergence is uniform since the improper integral converges uniformly. Furthermore each $F_n$ is continuous. This implies that the limit finction $F$ must be continuous (by another elementary theorem of real analysis).

Answer 2

Use the substitution $u^2=\sqrt{t}+4$ so that $u\to 2^+$ and the integral is reduced to $$\int_0^u\sqrt{1+\frac{(u^2-4)^2}{4\sqrt{x}}}\,dx$$ Let the integrand be denoted by $f(u, x) $ then we have $$0\leq f(u, x) - 1=\frac{(u^2-4)}{4\sqrt{x}}\cdot\dfrac{1}{1+\sqrt{1+\dfrac{(u^2-4)^2}{4\sqrt{x}}}}\leq \frac{(u^2-4)^2}{8\sqrt{x}}$$ Thus we get $$1\leq f(u, x) \leq 1+\frac{(u^2-4)^2}{8\sqrt{x}}$$ Integrating with respect to $x$ on $[0,u]$ we get $$u\leq \int_0^u f(u, x) \, dx\leq u+\frac{(u^2-4)^2\sqrt{u}}{4}$$ Letting $u\to 2^+$ we get desired limit as $2$.