Question

Estimate the limit. Picture below

Answer 1

Answer:

Hence, the limit of the expression:

$\lim_{x \to 3} \dfrac{x-3}{x^2-2x-3}$ is:

$\dfrac{1}{4}$

Step-by-step explanation:

We have to estimate the limit of:

$\lim_{x \to 3} \dfrac{x-3}{x^2-2x-3}$

We can also represent the denominator of the function in the limit as:

$x^2-2x-3=x^2-3x+x-3\\\\x^2-2x-3=x(x-3)+1(x-3)\\\\x^2-2x-3=(x+1)(x-3)$

Hence, we have to estimate the limit of:

$\lim_{x \to 3} \dfrac{x-3}{(x+1)(x-3)}\\\\= \lim_{x \to 3}\dfrac{1}{x+1}\\ \\=\dfrac{1}{3+1}\\\\=\dfrac{1}{4}$

Hence, the limit of the expression:

$\lim_{x \to 3} \dfrac{x-3}{x^2-2x-3}$ is:

$\dfrac{1}{4}$

Answer 2

Answer:

Choice C is correct answer.

Step-by-step explanation:

We have given expression.

$\lim_{x \to \ 3} x-3/x^{2} -2x-3$

We have to find the limit of function.

Simplifying the denominator, we have

x²-2x-3

Factoring the above expression, we have

(x-3)(x+1)

$\lim_{x \to \ 3} (x-3) / (x-3)(x+1)$

$\lim_{x \to \ 3} 1/x+1$

1 / 3+1

1/4

0.25

hence, $\lim_{x \to \ 3} x-3/x^{2} -2x-3$ = 0.25