Limits of Indeterminate Forms and Improper Integrals

**mturner07** · February 24th 2010, 07:28 PM

Hey guys, I have a few problems I need help with and answers checked.

Problem 1. Calculate each limit.

a. $\lim_{x \to 1}\frac{1 - x}{e^{x^3} - xe^x}$

My Solution

Form $\frac{0}{0}$ , Using L'Hopital's Rule. $= \lim_{x \to 1}\frac{-1}{3x^{2}e^{x^3} - e^x + xe^x} = \frac{-e}{3}$

b. $\lim_{x \to 0^+}\frac{e^{x^2} - 1}{\sin{x} - x}$

My Solution

Form $\frac{0}{0}$ , Using L'Hopital's Rule. = $\lim_{x \to 0^+}\frac{2xe^{x^2}}{\cos{x} - 1} = \frac{0}{0}$ , Using L'Hopital's Rule. $= \lim_{x \to 0^+}\frac{2e^{x^2} + 4x^{2}e^{x^2}}{-\sin{x}} = -\infty$

c. $\lim_{x \to 0}(\frac{1}{e^x - 1} - \frac{1}{x})$

My Solution

I'm having trouble with this one.

$= \lim_{x \to 0}\frac{x - e^x + 1}{xe^x - x} =$ Form $\frac{0}{0}$ , Using L'Hopital's Rule. $= \lim_{x \to 0}\frac{-2x + x^{2}e^x - e^{2x} - xe^{2x} + 2e^x - 1}{x^{2}e^{2x} - 2x^{2}e^x + x^2}$ ... This is also of the form $\frac{0}{0}$ . I've applied L'Hopital's Rule two more times and it just keeps getting nastier and nastier but is still keeping the form $\frac{0}{0}$ . Where am I going wrong?

d. $\lim_{x \to \infty}(1 + \frac{1}{x})^x$

My Solution

Let $y = (1 + \frac{1}{x})^x, ln y = ln(1 + \frac{1}{x})^x = xln(1 + \frac{1}{x})$

$\lim_{x \to \infty}ln y = \lim_{x \to \infty}xln(1 + \frac{1}{x}) = \lim_{x \to \infty}\frac{ln(1 + \frac{1}{x})}{\frac{1}{x}}$ = Form $\frac{0}{0}$ , Using L'Hopital's Rule. = (Shortening the simplifying steps due to me not wanting carpal tunnel syndrome.) $\lim_{x \to \infty}\frac{\frac{\frac{-1}{x^3}}{1 + \frac{1}{x}} - ln(1 + \frac{1}{x})(\frac{-1}{x^2})}{\frac{1}{x^2}}$ = $\lim_{x \to \infty}\frac{-x^4 + x^{2}ln(1 + \frac{1}{x})(x^3 + x^2)}{x^5 + x^4} = -1, lny \to -1, y \to e^{-1}$

Problem 2. Determine whether the improper integral converges, or diverges. In the case of convergence, give its value.

a. $\int^{e}_{0}lnx dx$

My Solution

$= \lim_{t \to 0^+}\int^{e}_{t} lnx dx$

u = lnx, du = $\frac{1}{x}$ dx, dv = dx, v = x

= $\lim_{t \to 0^+}xlnx - \int^{e}_{t} dx$

Here I don't know what I'm supposed to do. How am I supposed to get the limit if I have both x's and t's?

b. $\int^{\infty}_{2}xe^{-x} dx$

My Solution

Here I run into the same problem as in 2a.

Any and all help would be much appreciated. Thanks in advance.

**Random Variable** · February 24th 2010, 07:48 PM

$\lim_{b \to \infty} \int^{b}_{2} xe^{-x} \ dx = \lim_{b \to \infty} -xe^{-x} \Big|^{b}_{2} + \lim_{b \to \infty} \int^{b}_{2} e^{-x} \ dx$

$= \lim_{b \to \infty} (-xe^{-x} - e^{-x} \big|^{b}_2)$

$= \lim_{b \to \infty} (-be^{-b} - e^{-b} +2e^{-2}+e^{-2}) = 3e^{-2}$

because $\lim_{b \to \infty} be^{-b} = \lim_{b \to \infty} \frac{b}{e^{b}} = \lim_{b \to \infty} \frac{1}{e^{b}} = 0$

**Random Variable** · February 24th 2010, 08:08 PM

$\lim_{t \to 0+} \int^{e}_{t} \ln x \ dx = \lim_{t \to 0+} (x \ln x -x )\Big|^{e}_{t}$

$= \lim_{t \to 0+}(e \ln e - e - t \ln t + t)$

$= \lim_{t \to 0+}(e - e - t \ln t + t) = 0$

because $\lim_{t \to 0+} t \ln t = \lim_{t \to 0+} \frac{ \ln t}{1/t}= \frac{1/t}{-1/t^{2}} = \lim_{t \to 0+} -t = 0$

**Random Variable** · February 25th 2010, 04:53 AM

1a) It should be $\lim_{x \to 1}\frac{-1}{3x^{2}e^{x^3} - e^x \textcolor{red}{-} xe^x}$ .

1c) I'm sure what you did.

$\lim_{x \to 0} \frac{x-e^{x}+1}{xe^{x}-x} = \lim_{x \to 0} \frac{1-e^{x}}{e^{x}+xe^{x}-1}=$ $= \lim_{x \to 0} \frac{-e^{x}}{e^{x}+e^{x}+xe^{x}}= -\frac{1}{2}$

**mturner07** · February 25th 2010, 08:02 AM

Thank you very much for your help. Did I do problems 1b and 1c right?

**Aryth** · February 25th 2010, 08:31 AM

As far as I can tell, b is correct. For c I'll start with your consolidated form of it:

$\lim_{x \to 0} \frac{x - e^x + 1}{xe^x - x}$

Using L'Hopital's rule, we must differentiate the function:

$\frac{1 - e^x}{(e^x + xe^x) - 1}$

This is because we must differentiate the numerator and denominator separately, so that:

$\lim_{x \to 0} \frac{1 - e^x}{(e^x + xe^x) - 1} = \frac{0}{0}$

So, we use L'Hopital's rule again, getting:

$\frac{-e^x}{e^x + (e^x + xe^x)}$

So that:

$\lim_{x \to 0} \frac{-e^x}{e^x(2 + x)} = -\lim_{x \to 0} \frac{1}{2 + x} = -\frac{1}{2}$

So, Random Variable is right on b.

**mturner07** · February 25th 2010, 11:35 AM

I'm sorry, I meant is my solution to 1d correct (not 1b)?

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