Question

In Problems 7– 42, find each limit algebraically.

$\underset{x\to 2}{\lim }\frac{x^2-4}{x^2-2x}.$

Short answer

The answer is 2.

Step-by-step solution

Step. 1 Given Information

Firstly, we check whether the given function is in indeterminant form or not.

$f\left(x\right)=\frac{x^2-4}{x^2-2x}$

Since if we put $x=2$we get,

$\frac{2^2-4}{2^2-2\left(2\right)}=\frac{4-4}{4-4}=\frac{0}{0}$.

So given the limit is in 0/0 indeterminant form.

Step. 2 Try to find factor which make the function indeterminant.

Since we have both polynomial functions in the numerator and denominator, So let's factorize them,

$\frac{x^2-4}{x^2-2x}=\frac{x^2-2^2}{x\left(x-2\right)}=\frac{\left(x-2\right)\left(x+2\right)}{x\left(x-2\right)},$

cancelling $(x-2)$from both numerator and denominator we get,

$\frac{x+2}{x}$.

Now if we put $x=2$then it is not in indeterminant from. So, now we can directly put the limit in it.

Step. 3 Solving the limit

$$\begin{gathered} \underset{x\to 2}{\lim }\frac{x+2}{x}=\frac{2+2}{2} \\ =\frac{4}{2}=2. \end{gathered}$$