Limit of function with Square roots

Question

I have to evaluate the limit: $$\lim_{x\to 1}\frac{x^2-\sqrt{x}}{\sqrt{x}-1}$$

I have been instructed to apply the following identity ($ \forall \, a \neq -b$) $$a-b=\frac{a^2-b^2}{a+b}$$ I am unsure how to proceed. Can you please explain each step?

EDIT

My current approach. Please help me out to finish it, and correct me if I am wrong somewhere: $$(a-b)(a+b)=(a^2-b^2)$$ i.e $$\lim_{x\to 1}\frac{x^2-\sqrt{x}}{\sqrt{x}-1}.\lim_{x\to 1}\frac{x^2+\sqrt{x}}{x^2+\sqrt{x}}$$ = $$\lim_{x\to 1}\frac{x^4-x}{x^2\sqrt{x}+x-x^2-\sqrt{x}}$$ How must I proceed from here?

Answer 1

\begin{align} \lim_{x\to 1}\frac{x^2-\sqrt{x}}{\sqrt{x}-1}&=\lim_{x\to 1}\frac{\sqrt{x}[(\sqrt{x})^3-1]}{\sqrt{x}-1}\\ &=\lim_{x\to 1}\frac{\sqrt{x}(\sqrt{x}-1)(x+\sqrt{x}+1)}{\sqrt{x}-1}\\ &=\lim_{x\to 1}[\sqrt{x}(x+\sqrt{x}+1)]\\ &=3 \end{align}

If we have to use that identity, then we can use $\displaystyle \sqrt{x}-1=\frac{x-1}{\sqrt{x}+1}$.

\begin{align} \lim_{x\to 1}\frac{x^2-\sqrt{x}}{\sqrt{x}-1}&=\lim_{x\to 1}\frac{(x^2-\sqrt{x})(\sqrt{x}+1)}{x-1}\\ &=\lim_{x\to 1}\frac{x^2\sqrt{x}+x^2-x-\sqrt{x}}{x-1}\\ &=\lim_{x\to 1}\frac{(x^2-1)\sqrt{x}+x(x-1)}{x-1}\\ &=\lim_{x\to 1}\frac{(x-1)(x+1)\sqrt{x}+x(x-1)}{x-1}\\ &=\lim_{x\to 1}[(x+1)\sqrt{x}+x]\\ &=3 \end{align}

Answer 2

You can use L'hopital's rule:

$$\lim _{x\to 1} \frac{x^2-\sqrt x}{\sqrt x - 1}$$

Apply L'hopital's rule:

$$\lim _{x\to 1}\frac{2x-\frac{1}{2\sqrt x}}{\frac {1}{2 \sqrt x}} = \frac {1.5}{0.5} = 3$$

Answer 3

I expect that it is hinting that you multiply the top and bottom by ${\sqrt{x}+1}$.

$$\frac{x^2-\sqrt{x}}{\sqrt{x}-1} = \frac{(x^2-\sqrt{x})({\sqrt{x}+1})}{(\sqrt{x}-1)({\sqrt{x}+1})} = \frac{(x^2-\sqrt{x})({\sqrt{x}+1})}{{x-1}}$$

Answer 4

You can apply the substitution $\sqrt{x}=t$, so the limit becomes $$ \lim_{t\to1}\frac{t^4-t}{t-1}= \lim_{t\to1}\frac{t(t-1)(t^2+t+1)}{t-1} $$

With the hint $$ \lim_{x\to1}\frac{x^2-\sqrt{x}}{\sqrt{x}-1}= \lim_{x\to1}\frac{x^2-\sqrt{x}}{\sqrt{x}-1}\frac{\sqrt{x}+1}{\sqrt{x}+1}= \lim_{x\to1}\frac{x^2\sqrt{x}-x+x^2-\sqrt{x}}{x-1} $$ Now consider $$ x^2\sqrt{x}-x+x^2-\sqrt{x}= (x^2-1)\sqrt{x}+x(x-1)= (x+1)(x-1)\sqrt{x}+x(x-1) $$

Answer 5

HINT:

Set $\sqrt x-1=h,$

$\displaystyle \implies x^2-\sqrt x=(1+h)^4-(1+h)=3h+\binom42h^2+\binom43h^3+h^4$