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let $f(x)=\dfrac{x^2+x-2}{1-\sqrt{x}}$

How do I solve this limit? $$\lim_{x\to1}f(x)$$

I can replace the function with its content $$\lim_{x\to1}\dfrac{x^2+x-2}{1-\sqrt{x}}$$

Then rationalizing the denominator $$\lim_{x\to1}\dfrac{x^2+x-2}{1-\sqrt{x}}\cdot\dfrac{1+\sqrt{x}}{1+\sqrt{x}}$$

With $(a + b)(a - b) = a^2 - b^2$, I can remove the irrational denominator.

$$\lim_{x\to1}\dfrac{(x^2+x-2)(1+\sqrt{x})}{1-x}$$

I the multiply the two parenthesis $$\lim_{x\to1}\dfrac{\sqrt{x} \cdot x^2+\sqrt{x}\cdot x- \sqrt{x} \cdot 2 + x^2 + x - 2}{1-x}$$

I'm not sure where to continue to solve this limit.

Asaf Karagila
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Cedric Martens
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4 Answers4

10

Observe that $$ x^2 + x - 2 = (x-1)(x+2). $$ Therefore \begin{align} \lim_{x\to 1}\frac{(x^2+x-2)(1+\sqrt{x})}{1-x} &= \lim_{x\to 1} \frac{(x-1)(x+2)(1+\sqrt{x})}{1-x} \\ &= \lim_{x\to 1} \frac{-(1-x)(x+2)(1+\sqrt{x})}{1-x} \\ &= -\lim_{x\to 1} (x+2)(1+\sqrt{x}) \\ &= -(1+2)(1+\sqrt{1}) && \text{($\ast$)}\\ &= -6. \end{align} At step ($\ast$), we are using the fact that both factors are continuous at $x=1$ and that the product of continuous functions is continuous.

Xander Henderson
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7

Alternatively you can alwasy factor the hard way:

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

$\frac {x^{\frac 32}(\sqrt{x} - 1) + x^{\frac 32} + x - 2}{-\sqrt x + 1}=$

$-x^{\frac 32} + \frac {x(\sqrt x-1) + 2x - 2}{-\sqrt x+ 1}=$

$-x^{\frac 32} -x + \frac {2\sqrt x(\sqrt x -1) +2\sqrt x - 2}{-\sqrt x +1}=$

$-x^{\frac 32} - x - 2\sqrt{x}-2$

So $\lim\limits_{x\to 1} \frac {x^2 + x - 2}{-\sqrt x + 1}=\lim\limits_{x\to 1}(-x^{\frac 32} - x - 2\sqrt{x}-2) = -(1^{\frac 32}) - 1 - 2\sqrt 1 - 2= -1-1-2-2 = -6$.

Which is kind of difficult and hard and one ought to have seen and made use of $x^2 +x -2= (x+2)(x-1)$ and somehow noticed that $(x -1) = (\sqrt x+1)(\sqrt x -1)$ to make it easier.

But sometimes we don't see things and it's nice to know that brute force can work when we don't.

........

Or if we just let $y = \sqrt x$ then $\lim_{x\to 1} \frac {x^2 + x - 2}{1-\sqrt x }= \lim\limits_{y^2 \to 1;y > 0}\frac {y^4 + y^2 - 2}{1-y}$ and psychologically that seems easier. We just factor $y^4 + y^2 -2 = (y^2 + 2)(y^2 - 1) = (y^2+2)(y+1)(y-1)$ and the $y-1$s cancel out.

fleablood
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$$\lim_{x\to1}\dfrac{(x^2+x-2)(1+\sqrt{x})}{1-x}=\\ \lim_{x\to1}(1+\sqrt{x})\lim_{x\to1}\dfrac{(x^2+x-2)}{1-x}\\ =2\lim_{x\to1}\frac{(x+2)(x-1)}{1-x}=-6 $$

saulspatz
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2

$$\lim_{x\to 1}\frac{x^2+x-2}{1-\sqrt{x}}=\lim_{y\to 1}\frac{y^4+y^2-2}{1-y}=-\lim_{y\to 1}\frac{y^4+y^2-2}{y-1}$$

And that last limit is, by definition, the derivative $\frac{d}{dy}(y^4+y^2)=4y^3+2y$ at $y=1.$

So the result is $-6.$

Or you can just factor:

$$\frac{y^4+y^2-2}{y-1}=\frac{y^4-1}{y-1}+\frac{y^2-1}{y-1}=1+y+y^2+y^3 + 1+y.$$

Thomas Andrews
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