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Kipish [7]
3 years ago
13

Estimate the limit. Picture below

Mathematics
2 answers:
dezoksy [38]3 years ago
7 0

Answer:

what is it

Step-by-step explanation:

vova2212 [387]3 years ago
6 0

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}

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