Evaluating $\lim_{x\to 0} \frac{1-\sqrt{3x+1}}{2-\sqrt{5x+4}}$

Question

$$ \lim_{x\to 0} \frac{1-\sqrt{3x+1}}{2-\sqrt{5x+4}}$$

How do I solve this without using derivatives or integrals.

Answer 1

Both the numerator and the denominator are of the form $$ a-\sqrt{bx+a^2} $$ (with $a>0$). In the numerator we have $a=1$, $b=3$; in the denominator we have $a=2$, $b=5$.

Let's try \begin{align} \lim_{x\to0}\frac{a-\sqrt{bx+a^2}}{x} &=\lim_{x\to0}\frac{a-\sqrt{bx+a^2}}{x} \frac{a+\sqrt{bx+a^2}}{a+\sqrt{bx+a^2}}\\ &=\lim_{x\to0}\frac{-bx}{x(a+\sqrt{bx+a^2})}\\ &=\lim_{x\to0}\frac{-b}{a+\sqrt{bx+a^2}}\\ &=-\frac{b}{2a} \end{align} Thus your limit is $$ \lim_{x\to 0} \frac{1-\sqrt{3x+1}}{2-\sqrt{5x+4}}= \lim_{x\to 0} \frac{1-\sqrt{3x+1}}{x}\frac{x}{2-\sqrt{5x+4}}= \left(-\frac{3}{2}\right)\left(-\frac{4}{5}\right)=\frac{6}{5} $$

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When you'll learn Taylor expansions, you'll see that $$ \sqrt{1+kx}=1+\frac{1}{2}kx+o(x) $$ so $$ \frac{1-\sqrt{1+3x}}{2-\sqrt{4+5x}}= \frac{1-\sqrt{1+3x}}{2-2\sqrt{1+\frac{5}{4}x}}= \frac{1-(1+\frac{3}{2}x+o(x))}{2-2(1+\frac{5}{8}x+o(x))}= \frac{-\frac{3}{2}x+o(x)}{-\frac{5}{4}x+o(x)}= \frac{-\frac{3}{2}+o(1)}{-\frac{5}{4}+o(1)} $$ and so the limit is $$ \frac{-3/2}{-5/4}=\frac{6}{5} $$

Answer 2

Take a look at the following steps

$$\eqalign{ & L = \mathop {\lim }\limits_{x \to 0} {{1 - {{(3x + 1)}^{0.5}}} \over {2 - {{(5x + 4)}^{0.5}}}} = \mathop {\lim }\limits_{x \to 0} {{1 - {{(3x + 1)}^{0.5}}} \over {2 - {{(5x + 4)}^{0.5}}}}{{2 + {{(5x + 4)}^{0.5}}} \over {2 + {{(5x + 4)}^{0.5}}}}{{1 + {{(3x + 1)}^{0.5}}} \over {1 + {{(3x + 1)}^{0.5}}}} \cr & \,\,\,\, = \mathop {\lim }\limits_{x \to 0} {{1 - (3x + 1)} \over {4 - (5x + 4)}}{{2 + {{(5x + 4)}^{0.5}}} \over {1 + {{(3x + 1)}^{0.5}}}} \cr & \,\,\,\, = \mathop {\lim }\limits_{x \to 0} {{ - 3x} \over { - 5x}}{{2 + {{(5x + 4)}^{0.5}}} \over {1 + {{(3x + 1)}^{0.5}}}} \cr & \,\,\,\, = \mathop {\lim }\limits_{x \to 0} {{ - 3x} \over { - 5x}}\mathop {\lim }\limits_{x \to 0} {{2 + {{(5x + 4)}^{0.5}}} \over {1 + {{(3x + 1)}^{0.5}}}} \cr & \,\,\,\, = \left( {{3 \over 5}} \right)\left( {{4 \over 2}} \right) = {6 \over 5} \cr} $$