Homework Help: 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

@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.

21. Mar 2, 2016

Ssnow

22. Mar 3, 2016

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?

23. Mar 3, 2016

Samy_A

Do you know how to factor $a³-b³$?

24. Mar 3, 2016

gede

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

25. Mar 3, 2016

Samy_A

It should be:
$a³-b³=(a-b)(a²+ab+b²)$