# Math Help - Limit Marathon

1. Yep. That's it. Good job. It is easy if you spot it. Those are the fun ones.

2. Here is one, have fun with the simplicity

$\lim_{\psi\to{0^{+}}}\frac{\sin(\sqrt{\psi})\tan(e ^{10\psi}-1)}{\sqrt{15\psi^8-17\psi^3+64\psi}}$

3. Originally Posted by galactus
$\lim_{n\to\rightarrow{\infty}}\frac{1}{n}\sum_{j=1 }^{n}\sum_{k=1}^{n}\frac{j}{j^{2}+k^{2}}$
It would seem that this is 0 by a "power counting" argument. But I am having some trouble justifying it Mathematically.

-Dan

4. Originally Posted by Mathstud28
Here is one, have fun with the simplicity

$\lim_{\psi\to{0^{+}}}\frac{\sin(\sqrt{\psi})\tan(e ^{10\psi}-1)}{\sqrt{15\psi^8-17\psi^3+64\psi}}$
hint
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

5. Originally Posted by Mathstud28
Here is one, have fun with the simplicity

$\lim_{\psi\to{0^{+}}}\frac{\sin(\sqrt{\psi})\tan(e ^{10\psi}-1)}{\sqrt{15\psi^8-17\psi^3+64\psi}}$
This actually turns out to be really easy

Let $u(0)=0$

Then $\lim_{x\to{0}}\frac{\tan(u(x))}{u(x)}=1$

This can be done by making the sub with $\psi=u(x)$

$\therefore\tan(u(x))\sim{u(x)}$ if $u(0)$

and the sine case was listed earlier, the other case can be seen before which states that if

a polynomial approaches zero then it is ~ to its lowest powered x term

$\therefore\sqrt{15x^8-17x^3+64x}\sim{8\sqrt{x}}$

$\therefore\lim_{\psi\to{0^{+}}}\frac{\sin(\sqrt{\p si})\tan(e^{10\psi}-1)}{\sqrt{15\psi^8-17\psi^3+64\psi}}\sim\lim_{\psi\to{0^{+}}}\frac{\s qrt{\psi}(e^{10\psi}-1)}{8\sqrt{\psi}}=0$

6. Originally Posted by galactus
$\lim_{n\to\infty}\frac{1}{n}\sum_{j=1}^{n}\sum_{k= 1}^{n}\frac{j}{j^{2}+k^{2}}$
nice problem! the sum is a double Riemann sum of $f(x,y)=\frac{x}{x^2+y^2}$ on the unit square:

$\lim_{n\to\infty} \frac{1}{n}\sum_{j=1}^n \sum_{k=1}^n \frac{j}{j^2 + k^2}= \lim_{n\to\infty} \frac{1}{n^2} \sum_{j=1}^n\sum_{k=1}^n \frac{\frac{j}{n}}{(\frac{j}{n})^2 + (\frac{k}{n})^2}=\int_0^1 \int_0^1 \frac{x}{x^2 + y^2} \ dx \ dy$

$=\frac{1}{2}\int_0^1 (\ln(y^2 + 1) - 2\ln y) \ dy=\frac{\pi}{4} + \frac{\ln 2}{2}. \ \ \ \square$

7. Here are two simple ones

$\lim_{x\to{0}}\bigg[e^{\frac{5}{x}}-3x\bigg]^{\frac{x}{2}}$

and

$\lim_{x\to{0}}\frac{\sqrt{\cos(ax)}-\sqrt{\cos(bx)}}{x^2}$

8. Originally Posted by Mathstud28
Here are two simple ones

$\lim_{x\to{0}}\bigg[e^{\frac{5}{x}}-3x\bigg]^{\frac{x}{2}}$

and

$\lim_{x\to{0}}\frac{\sqrt{\cos(ax)}-\sqrt{\cos(bx)}}{x^2}$

Well I am sure there is a better way, but based one mine

Hint: make a substitution

Hint:conjugate

9. Originally Posted by Mathstud28
Here are two simple ones

$\lim_{x\to{0}}\bigg[e^{\frac{5}{x}}-3x\bigg]^{\frac{x}{2}}$

and

$\lim_{x\to{0}}\frac{\sqrt{\cos(ax)}-\sqrt{\cos(bx)}}{x^2}$
For the first one

$\lim_{x\to{0}}\bigg[e^{\frac{5}{x}}-3x\bigg]^{\frac{x}{2}}$

First I let

$L=\lim_{x\to\infty}\bigg[e^{\frac{5}{x}}-3x\bigg]^{\frac{x}{2}}\Rightarrow{L^2}=\lim_{x\to\infty}\b igg[e^{\frac{5}{x}}-3x\bigg]^{x}$

I then made the sub, let $\psi=\frac{1}{x}$

so as $x\to{0}\Rightarrow\psi\to\infty$

So we have that $x=\frac{1}{\psi}$

So by subbing we have

$L^2=\lim_{\psi\to\infty}\bigg[e^{5\psi}-\frac{3}{\psi}\bigg]^{\frac{1}{\psi}}$

Now there are a few ways this could have been done, but the easiest is the Root/Ratio laws test trick

$\lim_{\psi\to\infty}\bigg[e^{5\psi}-\frac{3}{\psi}\bigg]^{\frac{1}{\psi}}=\lim_{\psi\to\infty}\frac{e^5\cd ot{e^{5\psi}}-\frac{3}{\psi+1}}{e^{5\psi}-\frac{3}{\psi}}$

which then by termwise division gives

$\lim_{\psi\to\infty}\frac{e^5-\frac{3}{(\psi+1)e^{5\psi}}}{1-\frac{3}{\psi\cdot{e^{5\psi}}}}=e^5$

So $L^2=e^{5}\Rightarrow{L=e^{\frac{5}{2}}}$

For our second limit

multiplying by the conjugate gives

$\lim_{x\to{0}}\frac{\cos(ax)-\cos(bx)}{x^2\bigg[\sqrt{\cos(ax)}+\sqrt{\cos(bx)}\bigg]}$

Now by breaking it up we get

$\lim_{x\to{0}}\frac{\cos(ax)-\cos(bx)}{x^2}\cdot\lim_{x\to{0}}\frac{1}{\sqrt{\c os(ax)}+\sqrt{\cos(bx)}}=\lim_{x\to{0}}\frac{\cos( ax)-\cos(bx)}{2x^2}$

now seeing that

$\cos(ax)-\cos(bx)=\bigg[1-\frac{a^2x^2}{2}+\frac{a^4x^4}{24}-...\bigg]-\bigg[1-\frac{b^2x^2}{2}+\frac{b^4x^4}{24}-...\bigg]$

Grouping we get

$\cos(ax)-\cos(bx)=\bigg[1-1\bigg]+\frac{b^2-a^2}{2}x^2-\frac{b^4+a^2}{24}x^4+...\sim\frac{b^2-a^2}{2}x^2$

So then we have that

$\lim_{x\to{0}}\frac{\sqrt{\cos(ax)}-\sqrt{\cos(bx)}}{x^2}=\lim_{x\to{0}}\frac{\cos(ax)-\cos(bx)}{2x^2}\sim\lim_{x\to{0}}\frac{\frac{b^2-a^2}{2}x^2}{2x^2}=\frac{b^2-a^2}{4}$

10. Hi,

I've got a question for $\left(e^{\frac 5x}-3x\right)^{\frac x2}$

Assuming that x tends to 0 from the right (>0), $\frac 5x \to +\infty \implies e^{\frac 5x} \to +\infty$, while $3x \to 0$.

So we can neglect 3x with regard to $e^{\frac 5x}$

Therefore, the limit is $\left(e^{\frac 5x}\right)^{\frac x2}=e^{\frac 52}$

Also, I think you should be patient, we're not all in the same GMT and we're not as available as you are. So if no one answers within 2 hours, it's not a catastrophe...

11. Originally Posted by Moo
Hi,

I've got a question for $\left(e^{\frac 5x}-3x\right)^{\frac x2}$

Assuming that x tends to 0 from the right (>0), $\frac 5x \to +\infty \implies e^{\frac 5x} \to +\infty$, while $3x \to 0$.

So we can neglect 3x with regard to $e^{\frac 5x}$

Therefore, the limit is $\left(e^{\frac 5x}\right)^{\frac x2}=e^{\frac 52}$

Also, I think you should be patient, we're not all in the same GMT and we're not as available as you are. So if no one answers within 2 hours, it's not a catastrophe...
Haha thanks for pointing that out, I fixed that, if you look back I just literally stopped. I went and made a sandwich and when I came back I started on the second and didnt finish the first

As for thse second one

So if no one answers within 2 hours, it's not a catastrophe...
Yes it is

12. Ok, here are two more

This first one I did yesterday on another website, so if you have seen it, dont spoil it..not that its that hard

$\lim_{x\to{0}}\frac{a^{x^2}-\cos(x)}{\sin(x^2)}$

and the other is

$\lim_{n\to\infty}\bigg(\frac{3^n+5^n}{2^n+5^n}\big g)^{n(\sqrt{n+1}-\sqrt{n})}$

13. Originally Posted by Mathstud28
$\lim_{x\to{0}}\frac{\sqrt{\cos(ax)}-\sqrt{\cos(bx)}}{x^2}$
Just note that: $
\frac{{\sqrt {\cos \left( {ax} \right)} - \sqrt {\cos \left( {bx} \right)} }}
{{x^2 }} = - \tfrac{1}
{{2x}} \cdot \int_b^a {\tfrac{{\sin \left( {x \cdot y} \right)}}
{{\sqrt {\cos \left( {x \cdot y} \right)} }}dy}
$

$
\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {\cos \left( {ax} \right)} - \sqrt {\cos \left( {bx} \right)} }}
{{x^2 }} = - \tfrac{1}
{2} \cdot \mathop {\lim }\limits_{x \to 0} \int_b^a {\tfrac{{\sin \left( {x \cdot y} \right)}}
{{x \cdot \sqrt {\cos \left( {x \cdot y} \right)} }}dy} =
$
$
- \tfrac{1}
{2} \cdot \int_b^a {\mathop {\lim }\limits_{x \to 0} \tfrac{{\sin \left( {x \cdot y} \right)}}
{{x \cdot \sqrt {\cos \left( {x \cdot y} \right)} }}dy} = - \tfrac{1}
{2} \cdot \int_b^a {y \cdot dy}
$
(assuming uniform convergence)

Thus: $
\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {\cos \left( {ax} \right)} - \sqrt {\cos \left( {bx} \right)} }}
{{x^2 }} = \tfrac{{b^2 - a^2 }}
{4}
$

14. Originally Posted by Mathstud28
$\lim_{x\to{0}}\frac{a^{x^2}-\cos(x)}{\sin(x^2)}$
$L=\mathop {\lim }\limits_{x \to 0} \tfrac{{a^{x^2 } - \cos \left( x \right)}}
{{\sin \left( {x^2 } \right)}} = \mathop {\lim }\limits_{x \to 0} \tfrac{{a^{x^2 } - 1 + 1 - \cos \left( x \right)}}
{{\sin \left( {x^2 } \right)}}
$

Now: $
\sin \left( {x^2 } \right)\mathop \sim \limits_{x \to 0} x^2
$
and we'll split the limit in 2 parts since both are defined.

$
L = \mathop {\lim }\limits_{x \to 0} \tfrac{{a^{x^2 } - 1}}
{{x^2 }} + \mathop {\lim }\limits_{x \to 0} \tfrac{{1 - \cos \left( x \right)}}
{{x^2 }}
$

And: $
\left\{ \begin{gathered}
a^{x^2 } - 1 = e^{x^2 \ln \left( a \right)} - 1\mathop \sim \limits_{x \to 0} x^2 \ln \left( a \right) \hfill \\
1 - \cos \left( x \right)\mathop \sim \limits_{x \to 0} \tfrac{{x^2 }}
{2} \hfill \\
\end{gathered} \right.
$

So we get: $
L = \ln \left( a \right) + \tfrac{1}
{2}
$

15. Originally Posted by PaulRS
$L=\mathop {\lim }\limits_{x \to 0} \tfrac{{a^{x^2 } - \cos \left( x \right)}}
{{\sin \left( {x^2 } \right)}} = \mathop {\lim }\limits_{x \to 0} \tfrac{{a^{x^2 } - 1 + 1 - \cos \left( x \right)}}
{{\sin \left( {x^2 } \right)}}
$

Now: $
\sin \left( {x^2 } \right)\mathop \sim \limits_{x \to 0} x^2
$
and we'll split the limit in 2 parts since both are defined.

$
L = \mathop {\lim }\limits_{x \to 0} \tfrac{{a^{x^2 } - 1}}
{{x^2 }} + \mathop {\lim }\limits_{x \to 0} \tfrac{{1 - \cos \left( x \right)}}
{{x^2 }}
$

And: $
\left\{ \begin{gathered}
a^{x^2 } - 1 = e^{x^2 \ln \left( a \right)} - 1\mathop \sim \limits_{x \to 0} x^2 \ln \left( a \right) \hfill \\
1 - \cos \left( x \right)\mathop \sim \limits_{x \to 0} \tfrac{{x^2 }}
{2} \hfill \\
\end{gathered} \right.
$

So we get: $
L = \ln \left( a \right) + \tfrac{1}
{2}
$
I did it very similarly

$a^{x^2}-\cos(x)=e^{x^2\ln(a)}-\cos(x)=\bigg[1+\ln(a)x^2+\frac{\ln^2(a)x^4}{2}+...\bigg]-\bigg[1-\frac{x^2}{2}+\frac{x^4}{4!}-..\bigg]$

Combining we get

$a^{x^2}=\bigg[1-1\bigg]+\frac{2\ln(a)+1}{2}x^2+...\sim\bigg(\ln(a)+\frac{ 1}{2}\bigg)x^2$

and $\sin(x^2)=x^2-\frac{x^6}{3!}+\frac{x^{10}}{5!}-...\sim{x^2}$

$\therefore\lim_{x\to{0}}\frac{a^{x^2}-\cos(x)}{x^2}\sim\lim_{x\to{0}}\frac{\bigg(\ln(a)+ \frac{1}{2}\bigg)x^2}{x^2}=\ln(a)+\frac{1}{2}$

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