Limit Problem: Solving \lim_{x \to 2} \frac{\tan (2 - \sqrt{2x})}{x^2 - 2x}

[gede]

I have a limit problem. This is the problem:

\lim_{x \to 2} \frac{\tan (2 - \sqrt{2x})}{x^2 - 2x}

The solution is
\lim_{x \to 2} \frac{\tan (2 - \sqrt{2x})}{x^2 - 2x}
= \lim_{x - 2 \to 0} \frac{\tan [(2 - \sqrt{2x}) × \frac{2 + \sqrt{2x}}{2 + \sqrt{2x}} ]}{x(x - 2)}
= \lim_{x - 2 \to 0} \frac{\tan [ \frac{-2(x - 2)}{2 + \sqrt{2x}} ] }{x(x - 2)}

Let $p = x - 2$

and $2 + \sqrt{2x} = 2 + \sqrt{2⋅2} = 2 + \sqrt{4} = 2 + 2 = 4$

Then

\lim_{x - 2 \to 0} \frac{\tan [ \frac{-2(x - 2)}{2 + \sqrt{2x}} ] }{x(x - 2)}
= \lim_{p \to 0} \frac{\tan [-\frac{1}{2}p]}{2p}
= -\frac{1}{4}

Is this correct?

[NOTE: moved to this forum by mentor hence no homework template]

 

[Ssnow]

The result seems correct, but there are intermediate steps not well justified ...

 

[Ssnow]

you can directly write $\frac{\tan{(2-\sqrt{2x})}}{x^2-2x}=\frac{\tan{(2-\sqrt{2x})}(2-\sqrt{2x})}{(2-\sqrt{2x})(x^2-2x)}$, now $\lim_{x\rightarrow 2}\frac{\tan(2-\sqrt{2x})}{(2-\sqrt{2x})}=\lim_{t\rightarrow 0}\frac{\tan{t}}{t}=1$, with $t=2-\sqrt{2x}$ ... so what remains is $\lim_{x\rightarrow 2}\frac{(2-\sqrt{2x})}{x^2-2x}$ ...

 

[gede]

you can directly write $\frac{\tan{(2-\sqrt{2x})}}{x^2-2x}=\frac{\tan{(2-\sqrt{2x})}(2-\sqrt{2x})}{(2-\sqrt{2x})(x^2-2x)}$, now $\lim_{x\rightarrow 2}\frac{\tan(2-\sqrt{2x})}{(2-\sqrt{2x})}=\lim_{t\rightarrow 0}\frac{\tan{t}}{t}=1$, with $t=2-\sqrt{2x}$ ... so what remains is $\lim_{x\rightarrow 2}\frac{(2-\sqrt{2x})}{x^2-2x}$ ...

I don't understand. Is $\lim_{x \to 2} \frac{\tan (2 - \sqrt{2x})}{2 - \sqrt{2x}} = 1$?

 

[Ssnow]

yes

 

[CrazyNinja]

@Ssnow ... the answer is -1/4. Check your calculations again.

@gede... the answer you have received is correct. Instead of doing all this, why don't you directly use the L-Hospital's rule?

 

[gede]

@Ssnow ... the answer is -1/4. Check your calculations again.

@gede... the answer you have received is correct. Instead of doing all this, why don't you directly use the L-Hospital's rule?

How exactly this L'Hospital's rule like?

Please show me the work of it.

 

[Ssnow]

When in the limit you find indefinite forms as $\frac{\infty}{\infty},\frac{0}{0}$, you can derive the numerator and the denominator repeating the limit ... for the precise statement see: https://en.wikipedia.org/wiki/L'Hôpital's_rule

you can apply this to the limit $\lim_{x\rightarrow 2}\frac{2-\sqrt{2x}}{x^2-2x}$ obtaining the result ...

PS. The method more immediate but you must be familiar with the concept of derivation.

 

[CrazyNinja]

@Ssnow you mean differentiation, not derivation right?

 

[Ssnow]

@CrazyNinja yes, do the derivative ...

 

[gede]

Isn't $\lim \frac{\tan t}{t} = 1$ only valid if $t \to 0$?

So, why does $\lim_{t \to 2} \frac{\tan t}{t} = 1$?

 

[Ssnow]

Hi, I didn't write $\lim_{t\rightarrow 2}\frac{\tan{t}}{t}=1$, I wrote $\lim_{t\rightarrow 0}\frac{\tan{t}}{t}=1$. In fact let us $t=2-\sqrt{2x}$ then when $x\rightarrow 2$ we have that in the new variable $t\rightarrow 0$... (if you change variable in the limit also the interested point change but the limit remain the same ...)

 

[gede]

I still confuse. Could you please show me the page from inside of the bellow calculus textbook is about limit closest as above?

https://www.sendspace.com/file/9m9da0

 

[Ssnow]

In example 6 and 7 pag 106, limits are calculated with substitution methods

 

[gede]

In example 6 and 7 pag 106, limits are calculated with substitution methods

I'm still confuse. How do you get $-\frac{1}{4}$ ?

 

[Ssnow]

Ok, I show you the complete resolution that doesn't involve differentiation, we start with

\lim_{x \to 2} \frac{\tan (2 - \sqrt{2x})}{x^2 - 2x}

we multiply and divide by the factor $2-\sqrt{2x}$ so will be:

\lim_{x \to 2} \frac{\tan (2 - \sqrt{2x})}{2-\sqrt{2x}}\frac{2-\sqrt{2x}}{x^2 - 2x}

now the limit of a product is the product of limits.
We examine the first limit $\lim_{x\rightarrow 2}\frac{\tan (2 - \sqrt{2x})}{2-\sqrt{2x}}$ doing the substitution $t=2 - \sqrt{2x}$. In this case as $x\rightarrow 2$ we have $t\rightarrow 0$ and the limit is:

$\lim_{x\rightarrow 2}\frac{\tan (2 - \sqrt{2x})}{2-\sqrt{2x}}=\lim_{t\rightarrow 0}\frac{\tan{t}}{t}=1$

So the first limit is $1$. Now we shall examine the second $\lim_{x\rightarrow 2}\frac{2-\sqrt{2x}}{x^2 - 2x}$. We start adding and subtracting $x$, so

$\lim_{x\rightarrow 2}\frac{2-\sqrt{2x}}{x^2 - 2x}=\lim_{x\rightarrow 2}\frac{2-x+x-\sqrt{2x}}{x^2 - 2x}=$

splitting the ratio, we have:

$=\lim_{x\rightarrow 2}\frac{2-x}{x^2 - 2x}+\frac{x-\sqrt{2x}}{x^2 - 2x}=$

we simplify and rewrite $x^2 - 2x=(x-\sqrt{2x})(x+\sqrt{2x})$ (we used $A^2-B^2=(A-B)(A+B)$), so

$=\lim_{x\rightarrow 2}\frac{2-x}{x(x-2)}+\frac{x-\sqrt{2x}}{(x-\sqrt{2x})(x+\sqrt{2x})}=\lim_{x\rightarrow 2}\frac{-1}{x}+\frac{1}{(x+\sqrt{2x})}=-\frac{1}{2}+\frac{1}{4}=-\frac{1}{4}$

Now $1\cdot \left(-\frac{1}{4}\right)=-\frac{1}{4}$ and the total limit is $\frac{-1}{4}$.

 

[gede]

Can you show me from inside above calculus textbook about $\lim_{x \to 0} \frac{\tan x}{x} = 1$?

Is $\lim_{x \to 0} \frac{\cos x}{x} = 1$?

 

[Ssnow]

ok, in your book is proved that $\lim_{x\rightarrow 0}\frac{\sin{x}}{x}=1$, for the limit above we observe that:

$\lim_{x\rightarrow 0}\frac{\tan{x}}{x}=\lim_{x\rightarrow 0}\frac{\sin{x}}{x\cos{x}}=\lim_{x\rightarrow 0}\frac{\sin{x}}{x}\lim_{x\rightarrow 0}\frac{1}{\cos{x}}=1\cdot 1=1$

(the second is not true $\lim_{x\rightarrow 0}\frac{\cos{x}}{x}$ does not exists...)

 

[CrazyNinja]

@gede e I suggest you go through the basics of limits before attempting the question again. Feel free to ask any doubts. You can PM me or Ssnow.

 

[Mark44]

@gede I suggest you go through the basics of limits before attempting the question again.

Sounds like a good idea.

 

[Ssnow]

yes, I agree with @CrazyNinja

 

[gede]

How do you solve this limit problem?

\lim_{x \to 1} \frac{\tan (1 - x)}{x^3 - 1}

My solution is:

\lim_{x \to 1} \frac{\tan (1 - x)}{x^3 - 1}\frac{1 - x}{1 - x}
= \lim_{x \to 1} \frac{1 - x}{x^3 - 1}

What is the next solution?

 

[Samy_A]

How do you solve this limit problem?

\lim_{x \to 1} \frac{\tan (1 - x)}{x^3 - 1}

My solution is:

\lim_{x \to 1} \frac{\tan (1 - x)}{x^3 - 1}\frac{1 - x}{1 - x}
= \lim_{x \to 1} \frac{1 - x}{x^3 - 1}

What is the next solution?

Do you know how to factor $a³-b³$?

 

[gede]

Do you know how to factor $a³-b³$?

No. Please tell me. Is factoring $a^3 - b^3$ studied in algebra book?

 

[Samy_A]

No. Please tell me. Is factoring $a^3 - b^3$ studied in algebra book?

It should be:
$a³-b³=(a-b)(a²+ab+b²)$

 

[Ssnow]

The limit $\lim_{x\rightarrow 1}\frac{\tan{(1-x)}}{1-x}=1$ is correct, for the other $\lim_{x\rightarrow 1}\frac{1-x}{x^3-1}$ you must follow the advice of @Samy_A that involves the cubic difference $a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})$

 

[gede]

How to solve this limit?

\lim_{x \to 0} \frac{\sqrt{x} (x - 7)}{\sqrt{x} - \sqrt{7}}

This is what I get:

\lim_{x \to 0} \frac{\sqrt{x} (x - 7)}{\sqrt{x} - \sqrt{7}} \frac{\sqrt{x} + \sqrt{7}}{\sqrt{x} + \sqrt{7}}
= \lim_{x \to 0} \frac{\sqrt{x} (x - 7) (\sqrt{x} + \sqrt{7})}{(x - 7)}
= \lim_{x \to 0} \sqrt{x}(\sqrt{x} + \sqrt{7})

What is the next solution?

 

[Ssnow]

All it is correct but not necessary, I suggest you to put $x=0$ in the original limit, there are indefinite forms as $\frac{0}{0},\frac{\infty}{\infty}, 0\cdot \infty, \infty-\infty$ or not ?

 

[Ssnow]

I suggest you the same limit with $x\rightarrow 7$ instead $x\rightarrow 0$

 

[gede]

I don't understand.