# Is this limit solved correctly

Can you please check if i solved this limit correctly?

$\lim_{x \to 1} \frac{\sin{(1-x)}}{\sqrt{x}-1}$

To skip writing of code , I'll just explain what i did .. First, I multiplied the expression by $\frac{\sqrt{x}+1}{\sqrt{x}+1}$. Then i got $\frac{\sin{(1-x)} * (\sqrt{x}+1)}{x-1}$. Next i pulled out the minus in from of the brackets in denumerator so i got: $\frac{\sin{(1-x)} * (\sqrt{x}+1)}{-(1-x)}$. Since $\frac{\sin{(1-x)}}{1-x}$ is equal to 1, I am left with $-(\sqrt{x}+1)$. After exchanging x with 1 i get -2 as a result. Did i do this correctly?

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I think it is correct. – user46090 Oct 28 '12 at 10:21
It is correct for sure – Host-website-on-iPage Oct 28 '12 at 10:23
One note: $\frac{\sin{(1-x)}}{1-x} \ne{1}$ but $\lim\limits_{n\rightarrow\infty}\frac{\sin{(1-x)}}{1-x}=1$. – M. Strochyk Oct 28 '12 at 10:38
Yea, sorry, I'm new to limits so i forget to write it xD. Thanks – Transcendental Oct 28 '12 at 11:44

Applying L'Hospital (L'H), we get:

$$\lim_{x\to 1}\frac{\sin(1-x)}{\sqrt x-1}\stackrel{L'H}=\lim_{x\to 1}\frac{-\cos(1-x)}{\frac{1}{2\sqrt x}}=\frac{-1}{\frac{1}{2}}=-2$$

The above makes it sure you got the correct result in your development and that this last was very probably correct.

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I would have applied the rule but i didn't do derivatives yet and Wikipedia says that it has something to do with derivatives :D – Transcendental Oct 28 '12 at 11:47

$$\lim_{x \to 1} \frac{\sin{(1-x)}}{\sqrt{x}-1}$$

$$=\lim_{x \to 1} \frac{\sin (1-x)(\sqrt x+1)}{x-1}$$

Putting $h=1-x,$ so $h\to 0$ as $x\to 1,$ $$=-\lim_{h \to 0} \frac{\sin h(\sqrt {1-h}+1)}h$$

$$=-\lim_{h \to 0} \frac{\sin h}h \cdot \lim_{h \to 0}(\sqrt {1-h}+1)$$

$=-1\cdot 2=-2$

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