Answer:
Hence, the limit of the expression:
[tex]\lim_{x \to 3} \dfrac{x-3}{x^2-2x-3}[/tex] is:
[tex]\dfrac{1}{4}[/tex]
Step-by-step explanation:
We have to estimate the limit of:
[tex]\lim_{x \to 3} \dfrac{x-3}{x^2-2x-3}[/tex]
We can also represent the denominator of the function in the limit as:
[tex]x^2-2x-3=x^2-3x+x-3\\\\x^2-2x-3=x(x-3)+1(x-3)\\\\x^2-2x-3=(x+1)(x-3)[/tex]
Hence, we have to estimate the limit of:
[tex]\lim_{x \to 3} \dfrac{x-3}{(x+1)(x-3)}\\\\= \lim_{x \to 3}\dfrac{1}{x+1}\\ \\=\dfrac{1}{3+1}\\\\=\dfrac{1}{4}[/tex]
Hence, the limit of the expression:
[tex]\lim_{x \to 3} \dfrac{x-3}{x^2-2x-3}[/tex] is:
[tex]\dfrac{1}{4}[/tex]
