Answer:
Hence, the limit of the expression:[tex]\lim_{x \to 1} \dfrac{\dfrac{1}{x+2}-\dfrac{1}{3}}{x-1}[/tex] is:
[tex]\dfrac{-1}{9}[/tex]
Step-by-step explanation:
We are asked to estimate the limit of the expression:
[tex]\lim_{x \to 1} \dfrac{\dfrac{1}{x+2}-\dfrac{1}{3}}{x-1}[/tex]
We will simplify the expression by first taking the l.c.m of the terms in the numerator to obtain the expression as:
[tex]\dfrac{3-(x+2)}{3(x+2)}\\\\=\dfrac{3-x-2}{3(x+2)}\\\\=\dfrac{1-x}{3(x+2)}[/tex]
[tex]\lim_{x \to 1} \dfrac{1-x}{3(x+2)(x-1)}\\\\= \lim_{x \to 1} \dfrac{-(x-1)}{3(x+2)(x-1)}\\\\\\= \lim_{x \to 1} \dfrac{-1}{3(x+2)}[/tex]
since the same term in the numerator and denominator are cancelled out.
Now the limit of the function exist as the denominator is not equal to zero when x→1.
Hence,
[tex]\lim_{x \to 1} \dfrac{-1}{3(x+2)}\\\\=\dfrac{-1}{3(1+2)}\\\\=\dfrac{-1}{3\times 3}\\\\=\dfrac{-1}{9}[/tex]
Hence, the limit of the expression:[tex]\lim_{x \to 1} \dfrac{\dfrac{1}{x+2}-\dfrac{1}{3}}{x-1}[/tex] is:
[tex]\dfrac{-1}{9}[/tex]
