How to find $\text{lim}_{x \to 1} \dfrac{\sqrt{x} -1}{\sqrt{2x +3} - \sqrt{5}} $

Question

I'm trying to determine the limit $$ \text{lim}_{x \to 1} \dfrac{\sqrt{x} -1}{\sqrt{2x +3} - \sqrt{5}} $$

Already tried $$ \bigg( \dfrac{\sqrt{x} -1}{\sqrt{2x +3} - \sqrt{5}} \bigg)\times \bigg( \dfrac{\sqrt{2x +3} + \sqrt{5}}{\sqrt{2x +3} + \sqrt{5}} \bigg) $$ But it's still impossible.

Without using L'hôpital's rule

Answer 1

As you stated, the conjugate of the numerator of the fraction you have is $\sqrt x+1$ and the conjugate of the denominator is $\sqrt {2x+3}+\sqrt {5}\thinspace .$

If these conjugates are both used together, this leads us to the following solution :

$$ \begin{align}\frac{\sqrt{x} -1}{\sqrt{2x +3} - \sqrt{5}} &\cdot \frac{\sqrt{2x +3} + \sqrt{5}}{\sqrt{2x +3} + \sqrt{5}} \cdot \frac {\sqrt x+1}{\sqrt x+1} =\frac {\left(x-1\right)\left(\sqrt {2x+3}+\sqrt 5\right)}{2\left(\sqrt x+1\right)\left(x-1\right)}\thinspace .\end{align} $$

Note that the solution can also be reached using only the conjugate of the denominator of the fraction. However, we will need an additional mathematical operation such as factorization :

$$ \begin{align}\frac{\sqrt{x} -1}{\sqrt{2x +3} - \sqrt{5}}&=\frac{\sqrt{x} -1}{\sqrt{2x +3} - \sqrt{5}} \cdot \frac{\sqrt{2x +3} + \sqrt{5}}{\sqrt{2x +3} + \sqrt{5}}\\ &=\frac {\left(\sqrt x-1\right)\left(\sqrt {2x+3}+\sqrt {5}\right)}{2\left(x-1\right)}\\ &=\frac {\left(\sqrt x-1\right)\left(\sqrt {2x+3}+\sqrt {5}\right)}{2\left(\sqrt x-1\right)\left(\sqrt x+1\right)}\\ &=\frac {\sqrt {2x+3}+\sqrt {5}}{2\left(\sqrt x+1\right)}\thinspace .\end{align} $$

Thus, you have :

$$ \begin{align}\lim_{x\to 1}\frac{\sqrt{x} -1}{\sqrt{2x +3} - \sqrt{5}}&=\lim_{x\to 1}\frac {\sqrt {2x+3}+\sqrt {5}}{2\left(\sqrt x+1\right)}\\ &=\frac {\sqrt{5}}{2}\thinspace .\end{align} $$

Answer 2

Just another way: by substituting $x=1+y$ the limit becomes $$ \frac{1}{\sqrt{5}}\lim_{y\to0}\frac{\sqrt{1+y}-1}{\sqrt{1+\frac{2}{5}y}-1}= \frac{5}{2}\frac{1}{\sqrt{5}}\lim_{y\to0}\frac{\sqrt{1+y}-1}{y}\cdot\frac{\frac{2}{5}y}{\sqrt{1+\frac{2}{5}y}-1}=\frac{1}{2}\sqrt{5} $$

Answer 3

$\lim_{x\to 1} \frac{\sqrt{x}-1}{\sqrt{2x+3}-\sqrt{5}} \\ \text{Use L-Hopital now} \\ \lim_{x\to 1} \frac{\frac{1}{2\sqrt{x}}}{\frac{1\cdot2}{2\sqrt{2x+3}}} \\ \lim_{x\to 1} \frac{\sqrt{2x+3}}{2\sqrt{x}} = \frac{\sqrt{5}}{2} $

ask me any doubts if you've further.

Answer 4

Another way to see it, by the definition of derivative with

• $f(x)=\sqrt{x} \implies f'(x) = \frac1{2\sqrt x}$
• $g(x)=\sqrt{2x +3} \implies g'(x) = \frac1{\sqrt{2x +3}}$

we have

$$\lim_{x \to 1} \frac{\sqrt{x} -1}{\sqrt{2x +3} - \sqrt{5}}=\lim_{x \to 1} \frac{\sqrt{x} -1}{x-1}\frac{x-1}{\sqrt{2x +3} - \sqrt{5}}=\frac{f'(1)}{g'(1)}=\frac{\frac12}{\frac1{\sqrt 5}}=\frac{\sqrt 5}2$$