$\text { Find the limit using } \lim _ { x = 0 } \frac { \sin x } { x } = 1 \text {. }$
-$\lim _ { x \rightarrow 0 } \frac { x ^ { 2 } - 2 x + \sin x } { x }$
A) does not exist
B) 0
C) $- 1$
D) 1

Question

Quizplus · Accepted Answer

We can factor the numerator as:
$$x^2 - 2x + \sin{x} = x(x-2) + \sin{x}$$
Therefore, we can rewrite the limit as:
$$\lim_{x 	o 0} \frac{x(x-2) + \sin{x}}{x}$$
$$=\lim_{x 	o 0} \left( (x-2) + \frac{\sin{x}}{x} ight)$$
Using the limit given in the problem, we know that $\lim_{x	o 0} \frac{\sin{x}}{x} = 1.$
Therefore, the limit reduces to:
$$\lim_{x 	o 0} (x-2) + 1 = -1$$
Thus, the answer is (C), -1.

Find the limit using lim⁡x=0sin⁡xx=1. \text { Find the limit using } \lim _ { x = 0 } \frac { \sin x } { x } = 1 \text {. } Find the limit using limx=0​xsinx​=1.

$\text { Find the limit using } \lim _ { x = 0 } \frac { \sin x } { x } = 1 \text {. }$