Use series to evaluate the limit.
-$$\lim _ { x \rightarrow - } \frac { 1 + \ln \left( 1 + 2 x ^ { 2 } \right) - \cos 2 x } { x ^ { 2 } }$$
A) 2
B) 1
C) 4
D) 5

Question

Quizplus · Accepted Answer

To evaluate the given limit, we can use the series expansions for $\ln(1+y)$ and $\cos(x)$, where $y = 2x^2$ for the logarithm part, and simply $x$ for the cosine part. The series expansions are:$$\ln(1+y) = y - \frac{y^2}{2} + \frac{y^3}{3} - \cdots$$$$\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots$$Substituting $y = 2x^2$ into the series for $\ln(1+y)$, we get:$$\ln(1+2x^2) = 2x^2 - \frac{(2x^2)^2}{2} + \cdots = 2x^2 - 2x^4 + \cdots$$For $\cos(2x)$, using the series expansion:$$\cos(2x) = 1 - \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} - \cdots = 1 - 2x^2 + \frac{(2x)^4}{4!} - \cdots$$Now, substituting these into the original limit expression:$$\lim _ { x \rightarrow - } \frac { 1 + \ln \left( 1 + 2 x ^ { 2 } \right) - \cos 2 x } { x ^ { 2 } } = \lim _ { x \rightarrow - } \frac { 1 + (2x^2 - 2x^4 + \cdots) - (1 - 2x^2 + \frac{(2x)^4}{4!} - \cdots) } { x ^ { 2 } }$$Simplifying the expression by canceling terms and focusing on the leading terms:$$= \lim _ { x \rightarrow - } \frac { 2x^2 - 2x^2 + \cdots } { x ^ { 2 } } = \lim _ { x \rightarrow - } \frac { 2x^4 - 2x^4 + \cdots } { x ^ { 2 } }$$The leading terms cancel out, and we need to correct the simplification to accurately reflect the series expansion's contribution to the limit. The correct approach is to notice that after canceling the $1$ and $-1$ and combining like terms, the dominant term in the numerator that doesn't cancel out is $2x^2$, and the $x^2$ terms in the denominator will cancel with this, leaving us with:$$= \lim _ { x \rightarrow - } \frac { 4x^2 } { x ^ { 2 } } = 4$$Thus, the correct answer is $4$, which corresponds to choice C.

Use Series to Evaluate the Limit lim⁡x→−1+ln⁡(1+2x2)−cos⁡2xx2\lim _ { x \rightarrow - } \frac { 1 + \ln \left( 1 + 2 x ^ { 2 } \right) - \cos 2 x } { x ^ { 2 } }x→−lim​x21+ln(1+2x2)−cos2x​

Use Series to Evaluate the Limit $\lim _ { x \rightarrow - } \frac { 1 + \ln \left( 1 + 2 x ^ { 2 } \right) - \cos 2 x } { x ^ { 2 } }$