# Limit Question

1. Feb 28, 2016

### 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]

Last edited by a moderator: Feb 29, 2016
2. Feb 28, 2016

### Ssnow

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

3. Feb 28, 2016

### 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 $\lim_{x\rightarrow 2}\frac{(2-\sqrt{2x})}{x^2-2x}$ ...

4. Feb 28, 2016

### gede

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

5. Feb 29, 2016

### Ssnow

yes

6. Feb 29, 2016

### 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?

7. Feb 29, 2016

### gede

How exactly this L'Hospital's rule like?

Please show me the work of it.

8. Feb 29, 2016

### 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.

9. Feb 29, 2016

### CrazyNinja

@Ssnow you mean differentiation, not derivation right?

10. Feb 29, 2016

### Ssnow

11. Feb 29, 2016

### gede

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

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

12. Mar 1, 2016

### 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 ...)

13. Mar 1, 2016

### gede

14. Mar 1, 2016

### Ssnow

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

15. Mar 1, 2016

### gede

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

16. Mar 1, 2016

### 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 $\frac{-1}{4}$.

17. Mar 1, 2016

### gede

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

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

18. Mar 1, 2016

### 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...)

19. Mar 1, 2016

### 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.

20. Mar 1, 2016

### Staff: Mentor

Sounds like a good idea.

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