How to calculate limit in a quotient function

Question

How to calculate the limit of the following function:

$$\lim_{x \rightarrow 1}\frac{\sqrt[3]{2x^2-1} -\sqrt[2]{x} }{x-1}$$

I've tried to use the following formula: $(a^3-b^3)=(a-b)(a^2+ab+b^2)$ without any success. Is there anything that I'm missing?

I can't use L'Hôpital rule.

Answer 1

$$\frac{\sqrt[3]{2x^2-1} -\sqrt{x} }{x-1}=\frac{\sqrt[3]{2x^2-1}-1- (\sqrt{x}-1) }{x-1}=$$ $$=\frac{2(x+1)}{\sqrt[3]{(2x^2-1)^2}+\sqrt[3]{2x^2-1}+1}-\frac{1}{\sqrt{x}+1}\rightarrow\frac{4}{3}-\frac{1}{2}=\frac{5}{6}.$$

Answer 2

With $x=y+1$, you get $$\frac{\sqrt[3]{2x^2-1}-\sqrt{x}}{x-1} = \frac{\sqrt[3]{2(y+1)^2-1}-\sqrt{y+1}}{y}$$

You have to compute the limit of this quantity when $y$ tends to $0$.

Now $$\frac{\sqrt[3]{2(y+1)^2-1}-\sqrt{y+1}}{y} = \frac{\sqrt[3]{2y^2 + 4y +1}-\sqrt{y+1}}{y} = \frac{1+ \frac{1}{3}(2y^2+4y) + o(y)-1- \frac{1}{2}y + o(y)}{y} $$

i.e. $$\frac{\sqrt[3]{2(y+1)^2-1}-\sqrt{y+1}}{y} = \frac{5}{6} + o(1)$$

So the limit is equal to $$\frac{5}{6}$$

Answer 3

Notice $\lim_{x\to\:1}\left(\sqrt[3]{2x^2-1}-\sqrt{x}\right)=0$ and $\lim_{x\to\:1}\left(x-1\right)=0$

Thus, $\lim_{x\to\:1}\left(\frac{\sqrt[3]{2x^2-1}-\sqrt{x}}{x-1}\right)=\frac{0}{0}$. Which is also known as "indeterminate form". In such situations, we apply L'Hopital's rule, as follows:

By differentiating the numerator and denominator, we get: $$\lim_{x\to\:1}\left(\frac{\frac{4x}{3\left(2x^2-1\right)^{\frac{2}{3}}}-\frac{1}{2x^{\frac{1}{2}}}}{1}\right)$$

Tidy this up: $$\lim_{x\to\:1}\left(\frac{8x\sqrt{x}-3\left(2x^2-1\right)^{\frac{2}{3}}}{6\sqrt{x}\left(2x^2-1\right)^{\frac{2}{3}}}\right)$$

Set $x=1$:

$$\frac{8\cdot\:1\cdot\sqrt{1}-3\left(2\cdot\:1^2-1\right)^{\frac{2}{3}}}{6\sqrt{1}\left(2\cdot\:1^2-1\right)^{\frac{2}{3}}}$$

Evaluate the above, and hence the limit is $5/6$.

Answer 4

Hint: Let $a=\sqrt[3]{2x^2-1}$ and $b =\sqrt[2]{x}$, then $a^6 = (2x^2-1)^2=4x^4-4x^2+1$ and $b^6 = x^3$. Then $$a^6-b^6 = 4x^4-x^3-4x^2+1 $$ $$=(x-1)(4x^3+3x^2-x-1)$$

$$=(x-1)(4b^6+3b^4-b^2-1)$$

Thus $$\lim_{x \rightarrow 1}\frac{\sqrt[3]{2x^2-1} -\sqrt[2]{x} }{x-1}=\lim_{a,b \rightarrow 1}\frac{(a-b)(4b^6+3b^4-b^2-1)}{a^6-b^6}$$ $$=\lim_{a,b \rightarrow 1}\frac{4b^6+3b^4-b^2-1}{a^5+a^4b+a^3b^2+a^2b^3+ab^4+b^5}={5\over 6}$$

Answer 5

Basic limit laws and standard limit $$\lim_{x\to a} \frac{x^n-a^n}{x-a}=na^{n-1}\tag{1}$$ suffice here.

We have \begin{align*} L&=\lim_{x\to 1}\frac{\sqrt[3]{2x^2-1}-\sqrt{x}}{x-1}\\ &=\lim_{x\to 1}\frac{\sqrt[3]{2x^2-1}-1}{x-1}-\frac{x^{1/2}-1^{1/2}}{x-1}\\ &=\lim_{x\to 1}\frac{\sqrt[3]{2x^2-1}-1}{2x^2-1-1}\cdot 2(x+1)-\frac{1}{2}\\ &=4\lim_{t\to 1}\frac{t^{1/3}-1^{1/3}}{t-1}-\frac{1}{2}\text{ (putting }t=2x^2-1) \\ &=4\cdot\frac{1}{3}-\frac{1}{2}\\ &=\frac{5}{6} \end{align*}