Question

Estimate the limit. Picture below

Answer 1

Answer:

what is it

Step-by-step explanation:

Answer 2

Answer:

Hence, the limit of the expression:$\lim_{x \to 1} \dfrac{\dfrac{1}{x+2}-\dfrac{1}{3}}{x-1}$ is:

$\dfrac{-1}{9}$

Step-by-step explanation:

We are asked to estimate the limit of the expression:

$\lim_{x \to 1} \dfrac{\dfrac{1}{x+2}-\dfrac{1}{3}}{x-1}$

We will simplify the expression by first taking the l.c.m of the terms in the numerator to obtain the expression as:

$\dfrac{3-(x+2)}{3(x+2)}\\\\=\dfrac{3-x-2}{3(x+2)}\\\\=\dfrac{1-x}{3(x+2)}$

$\lim_{x \to 1} \dfrac{1-x}{3(x+2)(x-1)}\\\\= \lim_{x \to 1} \dfrac{-(x-1)}{3(x+2)(x-1)}\\\\\\= \lim_{x \to 1} \dfrac{-1}{3(x+2)}$

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,

$\lim_{x \to 1} \dfrac{-1}{3(x+2)}\\\\=\dfrac{-1}{3(1+2)}\\\\=\dfrac{-1}{3\times 3}\\\\=\dfrac{-1}{9}$

Hence, the limit of the expression:$\lim_{x \to 1} \dfrac{\dfrac{1}{x+2}-\dfrac{1}{3}}{x-1}$ is:

$\dfrac{-1}{9}$