<s>[INST] How to solve $\lim_{x\to1}=\frac{x^2+x-2}{1-\sqrt{x}}$? 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.
 [/INST] 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.
</s>