# Check my limits Solution

• Oct 6th 2009, 12:13 PM
mj.alawami
Check my limits Solution
1) $\displaystyle lim (x\rightarrow 1) \frac{\sqrt{x}-1}{x-1}$
$\displaystyle \frac{1}{\sqrt{x}+1} =\frac{1}{2}$

2)$\displaystyle lim(x\rightarrow -1) \frac{\sqrt{h+5}-2}{h+1}$
$\displaystyle ]\frac{1}{\sqrt{h+5}+2}=\frac{1}{4}$

3)$\displaystyle lim(x\rightarrow 1)\frac{^3\sqrt{x}-1}{x-1}$

$\displaystyle \frac{1}{^3\sqrt{x}+1}=\frac{1}{2}$

4)$\displaystyle lim(x\rightarrow 0) \frac{^3\sqrt{h+1}-1}{h}$

$\displaystyle \frac{1}{^3\sqrt{h+1}+1}=\frac{1}{2}$

If there are any mistake please correct them?

Thank you..
• Oct 6th 2009, 12:44 PM
Hello mj.alawami
Quote:

Originally Posted by mj.alawami
1) $\displaystyle lim (x\rightarrow 1) \frac{\sqrt{x}-1}{x-1}$
$\displaystyle \frac{1}{\sqrt{x}+1} =\frac{1}{2}$

2)$\displaystyle lim(x\rightarrow -1) \frac{\sqrt{h+5}-2}{h+1}$
$\displaystyle ]\frac{1}{\sqrt{h+5}+2}=\frac{1}{4}$

3)$\displaystyle lim(x\rightarrow 1)\frac{^3\sqrt{x}-1}{x-1}$

$\displaystyle \frac{1}{^3\sqrt{x}+1}=\frac{1}{2}$

4)$\displaystyle lim(x\rightarrow 0) \frac{^3\sqrt{h+1}-1}{h}$

$\displaystyle \frac{1}{^3\sqrt{h+1}+1}=\frac{1}{2}$

If there are any mistake please correct them?

Thank you..

The first two are fine, but there's a problem with #3 and #4. In these you need to use

$\displaystyle x^3-1 = (x-1)(x^2+x+1)$

So #3 simplifies to $\displaystyle \frac{1}{x^{\frac23}+x^{\frac13}+1}$

... and #4 is similar.

• Oct 7th 2009, 01:48 AM
mj.alawami
Quote:

Hello mj.alawamiThe first two are fine, but there's a problem with #3 and #4. In these you need to use

$\displaystyle x^3-1 = (x-1)(x^2+x+1)$

So #3 simplifies to $\displaystyle \frac{1}{x^{\frac23}+x^{\frac13}+1}$

... and #4 is similar.

Can you please show me the complete steps of rationalizing the numerator of #3 ?
• Oct 7th 2009, 01:58 AM
mr fantastic
Quote:

Originally Posted by mj.alawami
Can you please show me the complete steps of rationalizing the numerator of #3 ?

My advice for #3 is to make the substitution $\displaystyle x = t^3$ so that you have $\displaystyle \lim_{t \rightarrow 1} \frac{t - 1}{t^3 - 1}$. Now factorise the denominator, cancel the factor common to numerator and denominator and then take the limit.

I advise #4 be done in a similar way.
• Oct 7th 2009, 07:58 AM
Hello mj.alawami

Use Mr F's substitution, or you can do it directly, by writing $\displaystyle \sqrt[3]x$ as $\displaystyle x^{\frac13}$.

Note first what happens when you multiply out the brackets from $\displaystyle (x^{\frac13}-1)(x^{\frac23}+x^{\frac13}+1)$, using the laws of indices:

$\displaystyle (x^{\frac13}-1)\color{red}(x^{\frac23}+x^{\frac13}+1)\color{bla ck}= x^{\frac13}\color{red}(x^{\frac23}+x^{\frac13}+1)\ color{black}-1\color{red}(x^{\frac23}+x^{\frac13}+1)\color{blac k}=x + x^{\frac23}+x^{\frac13}-x^{\frac23}-x^{\frac13}-1=x-1$

So we can factorise the denominator: $\displaystyle x-1 =(x^{\frac13}-1)(x^{\frac23}+x^{\frac13}+1)$

So $\displaystyle \lim_{x\to 1}\frac{\sqrt[3]x-1}{x-1}=\lim_{x\to 1}\frac{x^{\frac13}-1}{(x^{\frac13}-1)(x^{\frac23}+x^{\frac13}+1)}$

$\displaystyle =\lim_{x\to 1}\frac{1}{x^{\frac23}+x^{\frac13}+1}$

$\displaystyle =\frac13$

$\displaystyle \lim_{h\to0}\frac{\sqrt[3]{h+1}-1}{h}$
$\displaystyle =\lim_{h\to0}\frac{(h+1)^{\frac13}-1}{(h+1)-1}$
...which is basically the same as #3, with $\displaystyle x$ replaced by $\displaystyle (h+1)$.