How to evaluate limits by simplifying algebraic expressions and using one-sided behavior

Key concept: These two limit questions look different on the surface, but they share the same habit: first check whether the expression can be simplified, then see whether the function approaches the target value from the correct side. That mindset saves a lot of time on basic calculus limits.

Step by step

(a) $\lim_{x\to 2}\frac{x^2-4}{x-2}$

Factor the numerator:

$x^2-4=(x-2)(x+2)$

So for $x\neq 2$,

$\frac{x^2-4}{x-2}=x+2$

Now the limit is easy:

$\lim_{x\to 2}(x+2)=4$

(b) $\lim_{x\to 1^-}(|x-1|-2)$

Because $x\to 1^-$, we have $x<1$, so $x-1<0$. That means

$|x-1|=-(x-1)=1-x$

Substitute that in:

$|x-1|-2=(1-x)-2=-x-1$

Now take the limit from the left:

$\lim_{x\to 1^-}(-x-1)=-2$

Final answers

• (a) $4$
• (b) $-2$

The main idea is exactly what you said: rewrite the expression into something equivalent near the point, then substitute or evaluate the one-sided form carefully.

Pitfall alert

A common slip is to cancel factors and then forget the domain restriction. You may simplify $\frac{x^2-4}{x-2}$ to $x+2$, but only for $x\neq 2$; that is fine for limits because limits care about nearby values, not the value at the point itself. For the absolute value limit, the usual mistake is treating $|x-1|$ as $x-1$ even when approaching from the left. One-sided limits need the correct sign case.

Try different conditions

If the first limit were one-sided, $\lim_{x\to 2^-}\frac{x^2-4}{x-2}$ and $\lim_{x\to 2^+}\frac{x^2-4}{x-2}$ would both still be $4$ because the simplified form $x+2$ works on both sides near $2$. For the second expression, if the limit were $\lim_{x\to 1^+}(|x-1|-2)$, then $|x-1|=x-1$ and the limit would be $-2$ as well. In this case the left and right limits match, so the two-sided limit would also be $-2$.

Further reading

factoring limits, absolute value one-sided limits, removable discontinuity