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2. (10∗3+15=45pts)(10∗3+15=45pts) Evaluate the following limit. (c) lim⁡x→2x−x+...

Oct 19, 2024

2. (10∗3+15=45pts)(10∗3+15=45pts) Evaluate the following limit. (c) lim⁡x→2x−x+2x−2limx→2x−2x− (d) lim⁡x→1[(x−1)2sin⁡1x−1]limx→1[(x−1)2sinx−11]

Generated Graph

Solution by Steps

step 1

To evaluate the limit

\lim_{x \to 2} \frac{x^2 + 2x - 2}{x - 2}

, we first substitute

x = 2

into the expression. This gives us

\frac{2^2 + 2(2) - 2}{2 - 2} = \frac{4 + 4 - 2}{0}

, which is an indeterminate form

step 2

We can simplify the expression by factoring the numerator:

x^2 + 2x - 2 = (x - 2)(x + 4)

. Thus, the limit becomes

\lim_{x \to 2} \frac{(x - 2)(x + 4)}{x - 2}

step 3

We can cancel the

(x - 2)

terms, leading to

\lim_{x \to 2} (x + 4)

. Now substituting

x = 2

gives us

2 + 4 = 6

step 4

Therefore, the limit evaluates to

6

step 5

For the second limit

\lim_{x \to 1} \frac{(x - 1)^2 \sin x}{x - 1}

, we can simplify it to

\lim_{x \to 1} (x - 1) \sin x

step 6

Substituting

x = 1

gives us

(1 - 1) \sin(1) = 0 \cdot \sin(1) = 0

Answer

The limit for part (c) is

6

and for part (d) is

0

Key Concept

Limits and indeterminate forms are crucial in calculus for evaluating expressions that do not initially yield a defined value.

Explanation

The process of simplifying expressions and applying limits allows us to resolve indeterminate forms and find the actual limit values.