The hard limits thread

OP

A level students only

Find

\displaystyle \lim_{x \rightarrow 0} \frac{\sqrt{x+a}-\sqrt{a}}{x}

with

a > 0

without appealing to L'Hopital's Theorem.

Reply 2

OP

Find

\displaystyle \lim_{x \rightarrow \infty} \sqrt{x^{2n} +x^n}-\sqrt{x^{2n}}

for

n \in \mathbb{N}

Reply 3

garfeeled

16

I thought this thread was going to be about something else entirely.

Reply 4

Lord of the Flies

19

This is a neat one; there is a quick, beautiful way of evaluating it - and also a terribly boring one.

\dfrac{1}{e^n}\left(1+\dfrac{n}{1!}+\dfrac{n^2}{2!}+\cdots+ \dfrac{n^n}{n!}\right)\to\; ?

Reply 5

FireGarden

14

Original post by atsruser

Find

\displaystyle \lim_{x \rightarrow \infty} \sqrt{x^{2n} +x^n}-\sqrt{x^{2n}}

for

n \in \mathbb{N}

\displaystyle \lim_{x \rightarrow \infty} \sqrt{x^{2n} +x^n}-\sqrt{x^{2n}}

for

n \in \mathbb{N}

Let

y=x^n

, so we write

\displaystyle \lim_{y \rightarrow \infty} \sqrt{y^2 +y}-y

Multiplying top and bottom by 1/y gives

\dfrac{ \sqrt{1+\frac{1}{y}}-1 }{\frac{1}{y} }

. Let

z=\frac{1}{y}

, so the limit becomes

\displaystyle\lim_{z\to 0}\dfrac{ \sqrt{1+z}-1}{z}

.

This is just the derivative of

\sqrt{1+z}

evaluated at 0. So the limit is

\frac{1}{2}

.

(edited 8 years ago)

Reply 6

OP

Original post by FireGarden

\displaystyle \lim_{x \rightarrow \infty} \sqrt{x^{2n} +x^n}-\sqrt{x^{2n}}

for

n \in \mathbb{N}

Let

y=x^n

, so we write

\displaystyle \lim_{y \rightarrow \infty} \sqrt{y^2 +y}-y

Multiplying top and bottom by 1/y gives

\dfrac{ \sqrt{1-\frac{1}{y}}-1 }{\frac{1}{y} }

. Let

z=\frac{1}{y}

, so the limit becomes

\displaystyle\lim_{z\to 0}\dfrac{ \sqrt{1-z}-1}{z}

.

This is just the derivative of

\sqrt{1-z}

evaluated at 0. So the limit is

-\frac{1}{2}

.

I think you went wrong when manipulating the y expression:

\sqrt{y^2+y}-y = \sqrt{y^2(1+\frac{1}{y})}-y = y(\sqrt{1+\frac{1}{y}}-1)=\frac{\sqrt{1+\frac{1}{y}}-1}{1/y}

Reply 7

FireGarden

14

Original post by atsruser

I think you went wrong when manipulating the y expression:

\sqrt{y^2+y}-y = \sqrt{y^2(1+\frac{1}{y})}-y = y(\sqrt{1+\frac{1}{y}}-1)=\frac{\sqrt{1+\frac{1}{y}}-1}{1/y}

Oh dear - I wrote the question down wrong. I guess my final answer also has a sign error, then. Shall edit..

Reply 8

OP

Original post by FireGarden

Oh dear - I wrote the question down wrong. I guess my final answer also has a sign error, then. Shall edit..

It does. Correct apart from that though. And a nice way to introduce a derivative too.

Reply 9

OP

Original post by Lord of the Flies

This is a neat one; there is a quick, beautiful way of evaluating it - and also a terribly boring one.

\dfrac{1}{e^n}\left(1+\dfrac{n}{1!}+\dfrac{n^2}{2!}+\cdots+ \dfrac{n^n}{n!}\right)\to\; ?

I had thought of approaching this by considering an iterated limit, but if we write:

f(k,n) = \dfrac{1}{e^k}\left(1+\dfrac{k}{1!}+\dfrac{k^2}{2!}+\cdots+ \dfrac{k^n}{n!}\right)

we seem to have:

\displaystyle \lim_{k \to \infty} \lim_{n \to \infty} f(k,n) =1

\displaystyle \lim_{n \to \infty} \lim_{k \to \infty} f(k,n) =0

so I'm not sure if we can gather anything from that.

Reply 10

OP

Find

\displaystyle \lim_{x \to \infty} x^{1/x}

Reply 11

Mpagtches

16

Make sure you don't go too far with this kind of stuff ... because then you'll have gone past the limit :wink:

Reply 12

Indeterminate

3

Original post by atsruser

Find

\displaystyle \lim_{x \to \infty} x^{1/x}

Use the chain rule for limits.

Let

u = \dfrac{1}{x}

and now observe that

\displaystyle \lim_{x \rightarrow \infty} \dfrac{1}{x} = 0

Hence

\displaystyle \lim_{x \rightarrow \infty} x^{1/x} = \lim_{u \rightarrow 0} \left(\dfrac{1}{u^u}\right) = 1

(edited 8 years ago)

Reply 13

Where's the hard sums thread? :eyeball:

Reply 14

Original post by atsruser

Find

\displaystyle \lim_{x \rightarrow \infty} \sqrt{x^{2n} +x^n}-\sqrt{x^{2n}}

for

n \in \mathbb{N}

Original post by FireGarden

...

Here's a different method that is good at dealing with differences of square roots appearing in limits in general:

You can rearrange the expression using the difference of two squares formula:

\displaystyle a^2-b^2 = (a+b)(a-b) \Rightarrow a-b = \dfrac{a^2-b^2}{a+b}

.

Letting

\displaystyle a=\sqrt{x^{2n} +x^n}

and

b=\sqrt{x^{2n} } \left( = x^n \right)

, we get:

\displaystyle \lim_{x \rightarrow \infty} \sqrt{x^{2n} +x^n}-\sqrt{x^{2n}}

\displaystyle = \lim_{x \rightarrow \infty} \dfrac{x^{2n} +x^n-x^{2n} }{ \sqrt{x^{2n} +x^n}+\sqrt{x^{2n}}}

\displaystyle = \lim_{x \rightarrow \infty} \dfrac{1}{ \sqrt{\dfrac{x^{2n} +x^n}{x^{2n} }} +1}

\displaystyle = \lim_{x \rightarrow \infty} \dfrac{1}{ \sqrt{1+\dfrac{1}{x^n} } +1}

\displaystyle = \dfrac{1}{\sqrt{1+0} +1} = \dfrac{1}{2}

.

(edited 8 years ago)

Reply 15

Indeterminate

3

Original post by Star-girl

Where's the hard sums thread? :eyeball:

Not a bad idea.

Feel free to create a thread for that too! :biggrin:

Reply 16

Original post by Indeterminate

Not a bad idea.

Feel free to create a thread for that too! :biggrin:

It was atsruser's idea, so I'd feel bad hijacking it. :colondollar:

Reply 17

OP

Original post by Star-girl

It was atsruser's idea, so I'd feel bad hijacking it. :colondollar:

I'll do it tomorrow - I've had an awful headache all day so haven't got round to it yet. I don't know many hard series though - I'm more interested in seeing other people's examples.

Reply 18

Original post by atsruser

I'll do it tomorrow - I've had an awful headache all day so haven't got round to it yet. I don't know many hard series though - I'm more interested in seeing other people's examples.

Dw - no rush (I was just teasing a little). Take your time and get better soon. :console:

Same. More interested in solving/trying to solve the problems instead of providing them.

Reply 19

OP