CoormaBak9

2021-01-13

Evaluate the following limits.
$\underset{\left(x,y\right)\to \left(1,2\right)}{lim}\frac{\sqrt{y}-\sqrt{x+1}}{y-x-1}$

Clara Reese

We have to find the limit:
$\underset{\left(x,y\right)\to \left(1,2\right)}{lim}\frac{\sqrt{y}-\sqrt{x+1}}{y-x-1}$
It is $\frac{0}{0}$ form, since there is square root in numerator so applying rationalization rule multiplying and dividing by $\sqrt{y}+\sqrt{x+1}$,
$\underset{\left(x,y\right)\to \left(1,2\right)}{lim}\frac{\sqrt{y}-\sqrt{x+1}}{\left(y-x-1\right)}=\underset{\left(x,y\right)\to \left(1,2\right)}{lim}\frac{\left(\sqrt{y}-\sqrt{x+1}\right)}{\left(y-x-1\right)}×\frac{\left(\sqrt{y}+\sqrt{x+1}\right)}{\left(\sqrt{y}+\sqrt{x+1}\right)}$
Since we know the identity, $\left(a-b\right)\left(a+b\right)={a}^{2}-{b}^{2}$
Simplifying further,
$\underset{\left(x,y\right)\to \left(1,2\right)}{lim}\frac{\left(\sqrt{y}{\right)}^{2}-\left(\sqrt{x+1}{\right)}^{2}}{\left(y-x-1\right)\left(\sqrt{y}+\sqrt{x+1}\right)}=\underset{\left(x,y\right)\to \left(1,2\right)}{lim}\frac{y-\left(x+1\right)}{\left(y-x-1\right)\left(\sqrt{y}+\sqrt{x+1}\right)}$
$=\underset{\left(x,y\right)\to \left(1,2\right)}{lim}\frac{\left(y-x-1\right)}{\left(y-x-1\right)\left(\sqrt{y}-\sqrt{x+1}\right)}$
$=\underset{\left(x,y\right)\to \left(1,2\right)}{lim}\frac{1}{\left(\sqrt{y}+\sqrt{x+1}\right)}$
$=\frac{1}{\left(\sqrt{2}+\sqrt{1+1}\right)}$
$=\frac{1}{\left(\sqrt{2}+\sqrt{2}\right)}$
$=\frac{1}{2\sqrt{2}}$
Hence, value of limits is $\frac{1}{2\sqrt{2}}$

Jeffrey Jordon