Question

$\lim_{x \to \infty} \frac{\int_{0}^{x^2} sin t^5 dt}{x^{12}}$

Answer 1

0

Answer 2

To evaluate the limit $\lim_{x \to \infty} \frac{\int_{0}^{x^2} \sin t^5 dt}{x^{12}}$, we first analyze the form of the limit as $x \to \infty$.

Let $N(x) = \int_{0}^{x^2} \sin t^5 dt$ and $D(x) = x^{12}$.

As $x \to \infty$, the denominator $D(x) = x^{12} \to \infty$.

For the numerator $N(x) = \int_{0}^{x^2} \sin t^5 dt$: As $x \to \infty$, the upper limit $x^2 \to \infty$. The integral $\int_{0}^{\infty} \sin t^5 dt$ is a type of Fresnel integral. To check its convergence, we can use the substitution $u = t^5$, so $t = u^{1/5}$ and $dt = \frac{1}{5} u^{-4/5} du$. The integral becomes $\int_{0}^{\infty} \sin u \cdot \frac{1}{5} u^{-4/5} du = \frac{1}{5} \int_{0}^{\infty} \frac{\sin u}{u^{4/5}} du$. This integral converges by Dirichlet's test, as $\frac{1}{u^{4/5}}$ is monotonically decreasing and tends to 0 as $u \to \infty$, and $\int \sin u du$ is bounded. Therefore, $\lim_{x \to \infty} \int_{0}^{x^2} \sin t^5 dt$ converges to a finite value.

Since the numerator approaches a finite value and the denominator approaches infinity, the limit is of the form $\frac{\text{finite value}}{\infty}$, which is 0.

Alternatively, if we assume the limit is of the indeterminate form $\frac{\infty}{\infty}$ (which is a common assumption for such problems in competitive exams, leading to the application of L'Hopital's rule even if not strictly applicable by definition), we can apply L'Hopital's rule.

Let $f(x) = \int_{0}^{x^2} \sin t^5 dt$ and $g(x) = x^{12}$. We need to find $f'(x)$ and $g'(x)$.

Using the Leibniz integral rule, $\frac{d}{dx} \int_{a(x)}^{b(x)} h(t) dt = h(b(x)) b'(x) - h(a(x)) a'(x)$: $f'(x) = \frac{d}{dx} \int_{0}^{x^2} \sin t^5 dt = \sin((x^2)^5) \cdot \frac{d}{dx}(x^2) - \sin(0^5) \cdot \frac{d}{dx}(0)$ $f'(x) = \sin(x^{10}) \cdot (2x) - 0$ $f'(x) = 2x \sin(x^{10})$

Now, find the derivative of the denominator: $g'(x) = \frac{d}{dx} (x^{12}) = 12x^{11}$

Applying L'Hopital's rule: $\lim_{x \to \infty} \frac{f(x)}{g(x)} = \lim_{x \to \infty} \frac{f'(x)}{g'(x)}$ $= \lim_{x \to \infty} \frac{2x \sin(x^{10})}{12x^{11}}$ $= \lim_{x \to \infty} \frac{\sin(x^{10})}{6x^{10}}$

Let $y = x^{10}$. As $x \to \infty$, $y \to \infty$. The limit becomes: $\lim_{y \to \infty} \frac{\sin y}{6y}$

We know that the sine function is bounded, i.e., $-1 \le \sin y \le 1$. Dividing by $6y$ (which is positive for $y \to \infty$): $\frac{-1}{6y} \le \frac{\sin y}{6y} \le \frac{1}{6y}$

As $y \to \infty$, $\frac{-1}{6y} \to 0$ and $\frac{1}{6y} \to 0$. By the Squeeze Theorem, $\lim_{y \to \infty} \frac{\sin y}{6y} = 0$.

Both methods yield the same result.