Concept:?L Hospitals Rule:In this method, first, we have to check whether the form of the function after substituting the limit is \(\frac{0}{0}\;or\frac{\infty }{\infty }\) .If \(\mathop {\lim }\limits_{x \to a} f\left( x \right) = \frac{0}{0}\;or\frac{\infty }{\infty }\) then we have to differentiate both the numerator and denominator with respect to x unless and until \(\mathop {\lim }\limits_{x \to a} f\left( x \right) = l \ne \frac{0}{0}\) where l is a finite valueIf f(x) is a rational function then factorize the numerator and denominator, cancel out the common factors and then evaluate the limit of the function f(x) by substituting x = a. Formula used:\(\frac{d}{dx}x^n=nx^{n-1}\)\(\frac{d}{dx}\sqrt x=\frac{1}{2\sqrt x}\)Calculation:Let y = \(\mathop {\lim }\limits_{x \to 1} \dfrac{x-1}{\sqrt{x+3}-2}\)We can see that, at x = 1, function give 0/0 from. Hence, Apply L' Hospital's rule? y = \(\mathop {\lim }\limits_{x \to 1} \dfrac{\frac{d}{dx}(x-1)}{\frac{d}{dx}(\sqrt{x+3}-2)}\)By using the formula discussed above? y = \(\mathop {\lim }\limits_{x \to 1} \dfrac{1}{{\frac{1}{2\sqrt{(x+3)}}}-0}\)By taking the limit? y = \(\frac{1}{\frac{1}{4}}\) = 4? \(\mathop {\lim }\limits_{x \to 1} \dfrac{x-1}{\sqrt{x+3}-2}\) = 4