1. Limits

$\lim_{x\rightarrow1}\frac{x^2-1}{\sqrt{3x+1}-\sqrt{5x-1}}$ ..How can i Solve this using Theroms??

2. Re: Limits

I'd suggest multiplying the limit by the conjugate of the bottom (multiply top and bottom by sqrt(3x+1)+sqrt(5x-1)). Work from there and see if you can simplify the limit a bit.

3. Re: Limits

Hello, srirahulan!

What "theorems" are you referring to?

$\lim_{x\to1}\frac{x^2-1}{\sqrt{3x+1}-\sqrt{5x-1}}$

Multiply top and bottom by the conjugate.

. . $\lim_{x\to1}\frac{x^2-1}{\sqrt{3x+1} - \sqrt{5x-1}} \cdot \frac{\sqrt{3x+1} + \sqrt{5x+1}}{\sqrt{3x+1} + \sqrt{5x-1}}$

. . . . $=\;\lim_{x\to1}\frac{(x^2-1)\left(\sqrt{3x+1} + \sqrt{5x-1}\;\!\right)}{(3x+1) - (5x-1)} \;=\;\lim_{x\to1}\frac{(x^2-1)\left(\sqrt{3x+1} + \sqrt{5x-1}\right)}{-2x + 2}$

. . . . $=\;\lim_{x\to1} \frac{(x-1)(x+1)\left(\sqrt{3x+1} + \sqrt{5x-1}\right)}{-2(x-1)} \;=\;\lim_{x\to1}\left[-\frac{(x+1)\left(\sqrt{3x+1} + \sqrt{5x-1}\right)}{2}\right]$

. . . . $=\;-\frac{2(\sqrt{4} + \sqrt{4})}{2} \;=\;-(2+2) \;=\;\boxed{-4}$