1 Answer

Peixe · Answer 1 · 2021-03-16T04:17:55+0000

Answer:

$\lim_(x \to 1)(x^2-1)/(sin(x-2))=0$

Step-by-step explanation:

Assuming the correct expression is to find the following limit:

$\lim_(x \to 1)(x^2-1)/(sin(x-2))$

Use the property the limit of the quotient is the quotient of the limits:

$\lim_(x \to 1)(x^2-1)/(sin(x-2))=(\lim_(x \to 1)x^2-1)/(\lim_(x \to 1)sin(x-2))$

Evaluate the numerator:

$(\lim_(x \to 1)x^2-1)/(\lim_(x \to 1)sin(x-2))=(1^2-1)/(\lim_(x \to1)sin(x-2))=(0)/(\lim_(x \to 1)sin(x-2)$

Evaluate the denominator:

$(0)/(\lim_(x \to1)sin(x-2))=0$

Please log in or register to add a comment.