Answer:
[tex]\lim_{x \to 1}\frac{\sqrt{x^{2}+3 }-2}{{x-1}}=\frac{1}{2}[/tex]
Step-by-step explanation:
[tex]\lim_{x \to 1}\frac{\sqrt{x^{2}+3 }-2}{{x-1}}[/tex]
if we put x = 1 directly in the epression above we will get an indeterminate form [tex]\frac{0}{0}[/tex], so instead we will use L’Hospital's Rule.
Take derivative of numerator and denominator separately and then apply the limit.
[tex]\frac{d}{d(x)}\sqrt{x^{2}+3 }-2 = \frac{x}{\sqrt{x^{2}+3 } }\\\frac{d}{d(x)} (x-1) = 1\\[/tex]
So now the expression becomes: By puttin x = 1 in the expression now we can find the solution
[tex]\lim_{x \to 1} \frac{\frac{x}{\sqrt{x^{2}+3 } }}{1}\\=\lim_{x \to 1}\frac{x}{\sqrt{x^{2}+3 }}\\=\frac{1}{\sqrt{1^{2}+3 }}\\=\frac{1}{\sqrt{1+3 }}\\=\frac{1}{\sqrt{4 }}\\=\frac{1}{2}[/tex]
