Question

limit X tends to 1 root x + 8 - root 8 x + 1 upon root 5 minus x minus root 7 x minus 3

Answer 1

Given:

limit X tends to 1 root x + 8 - root 8 x + 1 upon root 5 minus x minus root 7 x minus 3

To find:

limit X tends to 1 root x + 8 - root 8 x + 1 upon root 5 minus x minus root 7 x minus 3

Solution:

From given, we have,

limit X tends to 1 root x + 8 - root 8 x + 1 upon root 5 minus x minus root 7 x minus 3

$\lim _{x\to \:1}\left(\dfrac{\sqrt{x+8}-\sqrt{8x+1}}{\sqrt{5-x}-\sqrt{7x-3}}\right)$

Apply L-Hospital's rule

$=\lim _{x\to \:1}\left(\dfrac{\frac{1}{2\sqrt{x+8}}-\frac{4}{\sqrt{8x+1}}}{-\frac{1}{2\sqrt{5-x}}-\frac{7}{2\sqrt{7x-3}}}\right)$

$=\lim _{x\to \:1}\left(\dfrac{\left(\sqrt{8x+1}-8\sqrt{x+8}\right)\sqrt{-x+5}\sqrt{7x-3}}{\left(-\sqrt{7x-3}-7\sqrt{-x+5}\right)\sqrt{x+8}\sqrt{8x+1}}\right)$

put the value of x = 1

$=\dfrac{\left(\sqrt{8\cdot \:1+1}-8\sqrt{1+8}\right)\sqrt{-1+5}\sqrt{7\cdot \:1-3}}{\left(-\sqrt{7\cdot \:1-3}-7\sqrt{-1+5}\right)\sqrt{1+8}\sqrt{8\cdot \:1+1}}$

upon solving the above equation, we get,

x = 7/12

∴ limit X tends to 1 root x + 8 - root 8 x + 1 upon root 5 minus x minus root 7 x minus 3 = 7/12