Limit with integral - L'Hospital's rule

Question

Find $$ \lim_{x \rightarrow 0^+} \frac{\int_{0}^{x^2} (1 + \sin(t))^{1/t} dt}{x \sin(x)} $$ Let $$ F(t) \mbox{ such that } F'(t) = (1 + \sin(t))^{1/t} $$ Then we use L'Hospital's rule: $$ \lim_{x \rightarrow 0^+} \frac{\int_{0}^{x^2} (1 + \sin(t))^{1/t} dt}{x \sin(x)} = \frac{F'(x^2) - F'(0) }{\sin(x) + x \cos(x)} $$ but $$ F'(0) $$ is not defined (we have $1/t$ part as exponent)

Answer 1

Lebnitz rule is applied as follows:

$$\frac{d}{dx} \int_{f(x)}^{g(x)} h(t) dt$$ $$=h(f(x))\times f'(x) - h(g(x)) \times g'(x) \quad (1)$$

Now w.r.t to your question,

$$f(x) = 0$$ $$g(x) = x^2$$ $$h(t) = (1+\sin(t))^{1/t}$$

Now we'll prove that the following limit exists:

$$L = \lim_{t \to 0} h(t)$$ $$L = \lim_{t \to 0} (1+\sin(t))^{1/t}$$

For this we'll first evaluate the limit for $\log h(t)$.

$$\lim_{t \to 0} \log(h(t))$$ $$=\lim_{t \to 0}\frac{\log(1+\sin(t))}{t}$$

Now upon applying L'Hopital rule to the above limit we can establish that $L=e$

Therefore,

$$\lim_{t \to 0} (1+\sin(t))^{1/t} = e$$

Now substituting this back to equation (1), using $f$, $g$, $h$, mentioned above, we get:

$$\frac{d}{dx} \int_{0}^{x^2} (1+\sin(t))^{1/t}$$

$$=((1+\sin(x^2))^{1/x^2})\times 2x - (\lim_{t\to 0}(1+\sin(t))^{1/t})\times 0$$ $$=((1+\sin(x^2))^{1/x^2})\times 2x$$

Hope this helps.

Answer 2

Hint: $$\lim_{x \to 0^+}\frac{\int_0^{x^2} (1+\sin(t))^{1/t} \,dt}{x\sin(x)}=\lim_{x \to 0^+}\frac{2x (1+\sin(x^2))^{x^{-2}}}{\sin(x) + x\cos(x)}=2\lim_{x \to 0^+}\frac{(1+\sin(x^2))^{x^{-2}}}{\frac{\sin(x)}{x}+\cos(x)}.$$

Answer 3

If the integrand is denoted by $f(t) $ then $\lim_{t\to 0^{+}}f(t)=e$ (check this). If we define $f(0)=e$ then $f$ becomes continuous at $0$.

Next the expression under limit can be written as $$\frac{x}{\sin x} \cdot\frac{1}{x^2}\int_{0}^{x^2}f(t)\,dt$$ The first factor tends to $1$ and thus the desired limit is equal to $$\lim_{u\to 0^+}\frac{1}{u}\int_{0}^{u}f(t)\,dt$$ via substitution $u=x^2$. Since $f$ is continuous at $0$ by fundamental theorem of calculus the above limit is $f(0)=e$.