# Thread: Improper Integrals My Tutor Cannot Understand!!!

1. ## Improper Integrals My Tutor Cannot Understand!!!

Here are some problems I am having a hard time with. If anyone knows how to solve them It would be amazing. My tutor cannot even figure these out!!!

1. limx->2- sqrt(4-x^2)/x-2

2. limx->0+ x^3cotx

3. limx->infinity (1+1/x)^x

4. limx->2+ (1/x^2-4)-(sqrtx-1/x^2-4)

2. Are these limits or integrals? If they are integrals, what are their bounds?

3. Do you mean something like this:

$\displaystyle \int ^2_k \frac{\sqrt{4-x^2}}{x-2}$ where $\displaystyle k$ is a constant $\displaystyle -2\le k<2$

If this is what you mean then to solve this you have to treat it like a limit as such:

$\displaystyle \int ^2_k \frac{\sqrt{4-x^2}}{x-2}=\displaystyle \lim_{t\to 2^{-}} \int ^t_k \frac{\sqrt{4-x^2}}{x-2}$. where $\displaystyle t$ is any constant $\displaystyle k<t<2$. Then you would do your regular integration by letting $\displaystyle x=2 \sin u$ and then plugging in $\displaystyle t$ and $\displaystyle k$ into your $\displaystyle F(x)$. You end up with:

$\displaystyle \lim_{t\to 2^{-}} \left(\sqrt{4-t^2}-2\arcsin \left(\frac{t}{2}\right)\right)$ (I have omitted the part where you plug in $\displaystyle k$ since those will just be numbers thus not affected by the limit).

You solve the limit, you end up with $\displaystyle -\pi$.

Therefore, $\displaystyle \int ^2_k \frac{\sqrt{4-x^2}}{x-2}=-\pi-\left(\sqrt{4-k^2}-2\arcsin \left(\frac{k}{2}\right)\right)$

4. Perhaps you should recognize the third one as a familiar e limit.

$\displaystyle \lim_{x\rightarrow{\infty}}(1+\frac{1}{x})^{x}=e$

Let $\displaystyle t=\frac{1}{x}, \;\ x=\frac{1}{t}$

Then we get another fundamental limit:

$\displaystyle \lim_{t\rightarrow{0}}(1+t)^{\frac{1}{t}}=e$

To prove the latter, we can build on the differentiability of ln(x) at x=1.

$\displaystyle \frac{d}{dx}[ln(x)]=\frac{1}{x}=1$

$\displaystyle 1=\lim_{t\rightarrow{0}}\frac{ln(1+t)-ln(1)}{t}$

$\displaystyle =\lim_{t\rightarrow{0}}\frac{ln(1+t)}{t}$

$\displaystyle =\lim_{t\rightarrow{0}}ln(1+t)^{\frac{1}{t}}$

And it follows:

$\displaystyle e=e^{\lim_{t\rightarrow{0}}ln(1+t)^{\frac{1}{t}}}$

$\displaystyle e=\lim_{t\rightarrow{0}}e^{ln(1+t)^{\frac{1}{t}}}$

$\displaystyle =\lim_{t\rightarrow{0}}(1+t)^{\frac{1}{t}}$

5. Originally Posted by daciernoj
Here are some problems I am having a hard time with. If anyone knows how to solve them It would be amazing. My tutor cannot even figure these out!!!

1. limx->2- sqrt(4-x^2)/x-2

2. limx->0+ x^3cotx

3. limx->infinity (1+1/x)^x

4. limx->2+ (1/x^2-4)-(sqrtx-1/x^2-4)
4. is ambiguous and needs more brackets or latex typesetting.

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1. $\displaystyle = \lim_{x \rightarrow 2^{-}} \frac{\sqrt{(2-x)(2+x)}}{-(2-x)} = - \lim_{x \rightarrow 2^{-}}\frac{\sqrt{2+x}}{\sqrt{2-x}} = - \frac{4}{0^{+}} = -\infty$.

Those who are easily offended, please turn a blind eye to the second last bit

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2. $\displaystyle = \lim_{x \rightarrow 0^{+}} \frac{x^3 \, \cos x}{\sin x} = \lim_{x \rightarrow 0^{+}} \frac{x^3 \, \cos x}{x - \frac{x^3}{3!} + ....} = \lim_{x \rightarrow 0^{+}} \frac{x^2 \, \cos x}{1 - \frac{x^2}{3!} + ....} =$ you should be able to take it from here.