We are given with a limit and we need to find it's value so let's start !!!!
[tex]{\quad \qquad \blacktriangleright \blacktriangleright \displaystyle \sf \lim_{x\to 4}\dfrac{\sqrt{x}-\sqrt{3\sqrt{x}-2}}{x^{2}-16}}[/tex]
But , before starting , let's recall an identity which is the main key to answer this question
Consider The limit ;
[tex]{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{\sqrt{x}-\sqrt{3\sqrt{x}-2}}{x^{2}-16}}[/tex]
Now as directly putting the limit will lead to indeterminate form 0/0. So , Rationalizing the numerator i.e multiplying both numerator and denominator by the conjugate of numerator
[tex]{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{\sqrt{x}-\sqrt{3\sqrt{x}-2}}{x^{2}-16}\times \dfrac{\sqrt{x}+\sqrt{3\sqrt{x}-2}}{\sqrt{x}+\sqrt{3\sqrt{x}-2}}}[/tex]
[tex]{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}-\sqrt{3\sqrt{x}-2})(\sqrt{x}+\sqrt{3\sqrt{x}-2})}{(x^{2}-4^{2})(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}[/tex]
Using the above algebraic identity ;
[tex]{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x})^{2}-(\sqrt{3\sqrt{x}-2})^{2}}{(x-4)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}[/tex]
[tex]{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{x-(3\sqrt{x}-2)}{(x-4)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}[/tex]
[tex]{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{x-3\sqrt{x}+2}{\{(\sqrt{x})^{2}-2^{2}\}(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}[/tex]
[tex]{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{x-3\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}[/tex]
Now , here we need to eliminate (√x-2) from the denominator somehow , or the limit will again be indeterminate ,so if you think carefully as I thought after seeing the question i.e what if we add 4 and subtract 4 in numerator ? So let's try !
[tex]{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{x-3\sqrt{x}-2+4-4}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}[/tex]
[tex]{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(x-4)+2+4-3\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}[/tex]
Now , using the same above identity ;
[tex]{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}-2)(\sqrt{x}+2)+6-3\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}[/tex]
[tex]{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}-2)(\sqrt{x}+2)+3(2-\sqrt{x})}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}[/tex]
Now , take minus sign common in numerator from 2nd term , so that we can take (√x-2) common from both terms
[tex]{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}-2)(\sqrt{x}+2)-3(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}[/tex]
Now , take (√x-2) common in numerator ;
[tex]{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}-2)\{(\sqrt{x}+2)-3\}}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}[/tex]
Cancelling the radical that makes our limit again and again indeterminate ;
[tex]{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{\cancel{(\sqrt{x}-2)}\{(\sqrt{x}+2)-3\}}{\cancel{(\sqrt{x}-2)}(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}[/tex]
[tex]{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}+2-3)}{(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}[/tex]
[tex]{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}-1)}{(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}[/tex]
Now , putting the limit ;
[tex]{:\implies \quad \sf \dfrac{\sqrt{4}-1}{(\sqrt{4}+2)(4+4)(\sqrt{4}+\sqrt{3\sqrt{4}-2})}}[/tex]
[tex]{:\implies \quad \sf \dfrac{2-1}{(2+2)(4+4)(2+\sqrt{3\times 2-2})}}[/tex]
[tex]{:\implies \quad \sf \dfrac{1}{(4)(8)(2+\sqrt{6-2})}}[/tex]
[tex]{:\implies \quad \sf \dfrac{1}{(4)(8)(2+\sqrt{4})}}[/tex]
[tex]{:\implies \quad \sf \dfrac{1}{(4)(8)(2+2)}}[/tex]
[tex]{:\implies \quad \sf \dfrac{1}{(4)(8)(4)}}[/tex]
[tex]{:\implies \quad \sf \dfrac{1}{128}}[/tex]
[tex]{:\implies \quad \bf \therefore \underline{\underline{\displaystyle \bf \lim_{x\to 4}\dfrac{\sqrt{x}-\sqrt{3\sqrt{x}-2}}{x^{2}-16}=\dfrac{1}{128}}}}[/tex]
We can transform the limand into a proper rational expression by substitution.
Let y = √x. Then as x approaches 4, y will approach √4 = 2. So
[tex]\displaystyle \lim_{x\to4}\frac{\sqrt x - \sqrt{3 \sqrt x - 2}}{x^2 - 16} = \lim_{y\to2} \frac{y - \sqrt{3y-2}}{y^4 - 16}[/tex]
Now let z = √(3y - 2). Then as y approaches 2, z will approach √(3•2 - 2) = 2 as well. It follows that y = (z² + 2)/3, so that
[tex]\displaystyle \lim_{y\to2} \frac{y - \sqrt{3y-2}}{y^4-16} = \lim_{z\to2} \frac{\frac{z^2+2}3 - z}{\frac{(z^2+2)^4}{81}-16} \\\\ = \lim_{z\to2} \frac{27(z^2+2)-81z}{(z^2+2)^4 - 1296} \\\\ = 27 \lim_{z\to2} \frac{z^2 - 3z + 2}{z^8 + 8z^6 + 24z^4 + 32z^2 - 1280}[/tex]
Plugging z = 2 into the denominator returns a value of 0, which means z - 2 divides z⁸ + 8z⁶ + 24z⁴ + 32z² - 1280 exactly. Polynomial division shows that
[tex]\dfrac{z^8 + 8z^6 + 24z^4 + 32z^2 - 1280}{z-2} \\\\ = z^7+2z^6+12z^5+24z^4+72z^3+144z^2+320z+640[/tex]
and it's easy to see that the numerator is also divisible by z - 2, since
[tex]z^2 - 3z + 2 = (z - 1) (z - 2)[/tex]
So, we can eliminate the factor of z - 2 and we're left with
[tex]\displaystyle 27 \lim_{z\to2} \frac{z^2 - 3z + 2}{z^8 + 8z^6 + 24z^4 + 32z^2 - 1280} = 27 \lim_{z\to2}\frac{z-1}{z^7+\cdots+640}[/tex]
The remaining limand is continuous at z = 2, so we can evaluate the limit by direct substitution:
[tex]\displaystyle 27 \lim_{z\to2}\frac{z-1}{z^7+\cdots+640} = \frac{27}{3456} = \boxed{\frac1{128}}[/tex]
