Evaluating $\lim_{x \to 2}\frac{\sqrt{3x-5}-\sqrt[3]{x-1}}{x-2}$

Question

Find the limit: $$\lim_{x \to 2}\frac{\sqrt{3x-5}-\sqrt[3]{x-1}}{x-2}$$

I tried multiplying by the conjugate both numerator and denominator, but it wasn't helpful.

Any ideas?

Answer 1

Hint: $\displaystyle\lim_{x\to2}\frac{\sqrt{3x-5}-\sqrt[3]{x-1}}{x-2}=\lim_{x\to2}\frac{\sqrt{3x-5}-1}{x-2}-\lim_{x\to2}\frac{\sqrt[3]{x-1}-1}{x-2}.$

Answer 2

Let $y=x-2$, then $$L=.\lim_{y\rightarrow 0} \frac{(1+3y)^{1/2}-(1+y)^{1/3}}{y}$$ Use binomial approximation $(1+z)^p \approx 1+pz$ if $|z|<<1$. Then $$L=\lim_{y \rightarrow 0} \frac{(1+3y/2)-(1+y/3)}{y}= \lim_{y\rightarrow 0} \frac{7y}{6y}=7/6.$$

Since it is $0/0$ form you may also use L'Hospital rule, differentiate up and down separately:

$$L=\lim_{x\rightarrow 2} \frac{\sqrt{3x-5}-\sqrt{x-1}}{x-2}=\lim_{x\rightarrow 2}\left( \frac{3}{2\sqrt{3x-5}}-\frac{(x-1)^{-2/3}}{3} \right)=\frac{7}{6}.$$