Answer:
Hence, the limit of the expression:
is:
![(-1)/(9)](https://img.qammunity.org/2020/formulas/mathematics/high-school/2lj6wv8sruuvgtak95mp2541jw2witlmk4.png)
Explanation:
We are asked to estimate the limit of the expression:
![\lim_(x \to 1) ((1)/(x+2)-(1)/(3))/(x-1)](https://img.qammunity.org/2020/formulas/mathematics/high-school/s463xbwiusivdnts0p59g978e0hm78sswj.png)
We will simplify the expression by first taking the l.c.m of the terms in the numerator to obtain the expression as:
![(3-(x+2))/(3(x+2))\\\\=(3-x-2)/(3(x+2))\\\\=(1-x)/(3(x+2))](https://img.qammunity.org/2020/formulas/mathematics/high-school/m3gad7z6o0eucyza9wc29ykac1w4ocxmku.png)
![\lim_(x \to 1) (1-x)/(3(x+2)(x-1))\\\\= \lim_(x \to 1) (-(x-1))/(3(x+2)(x-1))\\\\\\= \lim_(x \to 1) (-1)/(3(x+2))](https://img.qammunity.org/2020/formulas/mathematics/high-school/oc3ygxn06coc75srnsz16ecc32xv449ado.png)
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) (-1)/(3(x+2))\\\\=(-1)/(3(1+2))\\\\=(-1)/(3* 3)\\\\=(-1)/(9)](https://img.qammunity.org/2020/formulas/mathematics/high-school/casatn6tc3x0ckc4t3mr1fzryxczjuw839.png)
Hence, the limit of the expression:
is:
![(-1)/(9)](https://img.qammunity.org/2020/formulas/mathematics/high-school/2lj6wv8sruuvgtak95mp2541jw2witlmk4.png)