Find the limit of the function

[Sam Donovan]

Homework Statement

Find the limit of the function ( attached)

Homework Equations

The Attempt at a Solution

Is the limit infinity. Found the integral as (-1/(2t))*e^-2t from 0 to 0.
Isn't the integral of a function that is 0 to 0, 0. The lim of 1/x approaching zero infinity.

 

[LCKurtz]

Homework Statement

Find the limit of the function ( attached)

Homework Equations

The Attempt at a Solution

Is the limit infinity. Found the integral as (-1/(2t))*e^-2t from 0 to 0.
Isn't the integral of a function that is 0 to 0, 0. The lim of 1/x approaching zero infinity.

Apparently you are just plugging in $x=0$, and getting $0$ for the integral and $0$ for the denominator. In other words, a $\frac 0 0$ form. You can't draw any conclusion from that. But you likely have studied techniques to handle $\frac 0 0$ forms, right?

 

[arpon]

Look, if you put $x=0$, you have $\frac{0}{0}$ form [$\int_0^x e^{-t^2}dt$ is zero when $x=0$]. So in that case, you cannot conclude that the limit is infinity.
Now, look at the figure:

If $x\rightarrow 0$, what will be $\frac{\int_0^x e^{-t^2}dt}{x}=\frac{Area}{x}$ ? [Use trapezoidal approximation of area under the curve, that is, if $x$ is small enough, the area under the curve can be approximated by the area of the trapezoid $ABCD$]

 

[LCKurtz]

If $x\rightarrow 0$, what will be $\frac{\int_0^x e^{-2t}dt}{x}=\frac{Area}{x}$ ?

The integrand is $e^{-t^2}$, not $e^{-2t}$ so the area can't be calculated directly.

 

[arpon]

The integrand is $e^{-t^2}$, not $e^{-2t}$ so the area can't be calculated directly.

Thanks for pointing out the mistake. I did not mean to analytically calculate the area, I meant to use the trapezoidal approximation for the area.

 

[SammyS]

Homework Statement

Find the limit of the function ( attached)

Hello Sam Donovan. Welcome to PF !

It's a lot easier to follow the discussion with the following being visible.

 

[Ray Vickson]

Thanks for pointing out the mistake. I did not mean to analytically calculate the area, I meant to use the trapezoidal approximation for the area.

If you look at the graph of the function $e^{-t^2}$ you will see that the trapezoidal area $T(x)$ underestimates the exact area $A(x)$, so
$$\frac{T(x)}{x} < \frac{A(x)}{x}.$$
Thus, you can conclude that $\lim_{t \to 0} (1/x) \int_0^x f(t) \, dt \geq \lim_{x \to 0} T(x)/x$, and so get a lower bound on your desired limit. How can you get an upper bound as well, in order to extract an exact value?

 

[arpon]

If you look at the graph of the function $e^{-t^2}$ you will see that the trapezoidal area $T(x)$ underestimates the exact area $A(x)$, so
$$\frac{T(x)}{x} < \frac{A(x)}{x}.$$
Thus, you can conclude that $\lim_{t \to 0} (1/x) \int_0^x f(t) \, dt \geq \lim_{x \to 0} T(x)/x$, and so get a lower bound on your desired limit. How can you get an upper bound as well, in order to extract an exact value?

$$e^{-t^2}\leq1$$
$$\int^x_0e^{-t^2}\,dt\leq\int^x_0\,dt$$
$$\frac{\int^x_0e^{-t^2}\,dt}{x}\leq\frac{\int^x_0\,dt}{x}$$

 

[Ray Vickson]

$$e^{-t^2}\leq1$$
$$\int^x_0e^{-t^2}\,dt\leq\int^x_0\,dt$$
$$\frac{\int^x_0e^{-t^2}\,dt}{x}\leq\frac{\int^x_0\,dt}{x}$$

Right!

An even easier (but looser) lower bound than the trapezoidal one is obtainable from the fact that for $t \in (0,x)$ we have $e^{-t^2}> e^{-x^2}$, so that $\int_0^x e^{-t^2} \, dt > \int_0^t e^{-x^2} \, dt = x e^{-x^2}$. Besides, for some other, similar, functions, the trapezoidal estimate produces an upper bound, rather than a lower bound, so you are left unable to make a safe conclusion about the limit; for example because you do not have both a lower and upper bound. For example, consider $\int_0^x e^{-\sqrt{t}} \, dt$ for small $x > 0$.