Limits by rationalization

As $x\to 0$ the numerator tends to $\sqrt{1}-1=0$ and the denominator tends to $0$, therefore we have the indeterminate form $\frac{0}{0}$. Multiply numerator and denominator by $\sqrt{1+x}+1$, that is, rationalize: in the numerator the product $(a-b)(a+b)=a^2-b^2$ appears with $a=\sqrt{1+x}$ and $b=1$, which gives $x$. The factor $x$ cancels since in the limit $x$ is different from $0$. The remaining function is continuous at $0$, hence the limit is its value at $0$, that is, $\frac{1}{2}$.

As $x\to 4$ the numerator tends to $\sqrt{4}-2=0$ and the denominator tends to $0$, therefore we have the indeterminate form $\frac{0}{0}$. Multiply numerator and denominator by $\sqrt{x}+2$: in the numerator the product $(a-b)(a+b)=a^2-b^2$ appears with $a=\sqrt{x}$ and $b=2$, which gives $x-4$. The factor $x-4$ cancels since in the limit $x\ne 4$. The function $\frac{1}{\sqrt{x}+2}$ is continuous at $4$, hence the limit is its value at $4$, that is, $\frac{1}{\sqrt{4}+2}=\frac{1}{4}$.

As $x\to 1$ the numerator tends to $\sqrt{4}-2=0$ and the denominator tends to $\sqrt{1}-1=0$, therefore we have the indeterminate form $\frac{0}{0}$. Roots appear both in the numerator and in the denominator, hence one rationalizes twice: multiply by $\sqrt{x+3}+2$, which turns the numerator into $(x+3)-4=x-1$, and by $\sqrt{x}+1$, which turns the denominator into $x-1$. In both cases the product $(a-b)(a+b)=a^2-b^2$ is used. The common factor $x-1$ cancels since in the limit $x\ne 1$. The remaining function is continuous at $1$, hence the limit is its value at $1$, that is, $\frac{2}{4}=\frac{1}{2}$.