Find Limits Using Series

To find the limit \( \lim_{x \to 0}\dfrac{1 - \dfrac{1}{2}x^2 - \cos\left(\dfrac{x}{1 - x^2}\right)}{x^4} \), we use Taylor series expansions.

Expand the argument of the cosine function:

\[ \dfrac{x}{1 - x^2} = x \left(1 + x^2 + x^4 + \cdots \right) = x + x^3 + x^5 + \cdots \]

For small \( x \), we approximate this as \( \theta = x + x^3 \).

Expand \( \cos(\theta) \) using the Taylor series:

\[ \cos(\theta) = \cos(x + x^3) \approx 1 - \dfrac{\theta^2}{2} + \dfrac{\theta^4}{24} - \cdots \]

Calculate \( \theta^2 \) and \( \theta^4 \) up to \( x^4 \) terms:

\[ \theta^2 = (x + x^3)^2 = x^2 + 2x^4 + \cdots \] \[ \theta^4 = (x + x^3)^4 = x^4 + \cdots \]

Substitute these into the cosine expansion:

\[ \cos(x + x^3) \approx 1 - \dfrac{x^2 + 2x^4}{2} + \dfrac{x^4}{24} = 1 - \dfrac{x^2}{2} - x^4 + \dfrac{x^4}{24} \]

Simplify the terms:

\[ \cos(x + x^3) \approx 1 - \dfrac{x^2}{2} - \dfrac{23x^4}{24} \]

Substitute this back into the numerator of the original expression:

\[ 1 - \dfrac{1}{2}x^2 - \cos\left(\dfrac{x}{1 - x^2}\right) \approx 1 - \dfrac{1}{2}x^2 - \left(1 - \dfrac{x^2}{2} - \dfrac{23x^4}{24}\right) \]

Simplify:

\[ 1 - \dfrac{1}{2}x^2 - 1 + \dfrac{1}{2}x^2 + \dfrac{23x^4}{24} = \dfrac{23x^4}{24} \]

Divide the simplified numerator by the denominator \( x^4 \) and simplify:

\[ \dfrac{\dfrac{23x^4}{24}}{x^4} = \dfrac{23}{24} \]

Thus, the limit is:

\[ \boxed{ \lim_{x \to 0}\dfrac{1 - \dfrac{1}{2}x^2 - \cos\left(\dfrac{x}{1 - x^2}\right)}{x^4} = \dfrac{23}{24} } \]

To find the limit \( \lim_{x \to 0} \left( \dfrac{1}{\sin^2 x} + \dfrac{1}{\tan^2 x} - \dfrac{2}{x^2} \right) \), we start by using trigonometric identities and Taylor series expansions.

Rewrite the expression using trigonometric identities:

\[ \dfrac{1}{\tan^2 x} = \dfrac{\cos^2 x}{\sin^2 x} \]

Thus, the expression becomes:

\[ \dfrac{1}{\sin^2 x} + \dfrac{\cos^2 x}{\sin^2 x} - \dfrac{2}{x^2} = \dfrac{1 + \cos^2 x}{\sin^2 x} - \dfrac{2}{x^2} \]

Expand \( \cos x \) and \( \sin x \) using Taylor series:

\[ \cos x \approx 1 - \dfrac{x^2}{2} + \dfrac{x^4}{24} \] \[ \sin x \approx x - \dfrac{x^3}{6} \]

Squaring the above expansions:

\[ \cos^2 x \approx \left(1 - \dfrac{x^2}{2} + \dfrac{x^4}{24}\right)^2 \approx 1 - x^2 + \dfrac{x^4}{3} \] \[ \sin^2 x \approx \left(x - \dfrac{x^3}{6}\right)^2 \approx x^2 - \dfrac{x^4}{3} \]

Substitute these expansions into the expression:

\[ 1 + \cos^2 x \approx 2 - x^2 + \dfrac{x^4}{3} \] \[ \dfrac{1 + \cos^2 x}{\sin^2 x} \approx \dfrac{2 - x^2 + \dfrac{x^4}{3}}{x^2 - \dfrac{x^4}{3}} = \dfrac{2 - x^2 + \dfrac{x^4}{3}}{x^2 \left(1 - \dfrac{x^2}{3}\right)} \]

Simplifying:

\[ \dfrac{2 - x^2 + \dfrac{x^4}{3}}{x^2 \left(1 - \dfrac{x^2}{3}\right)} \approx \left( \dfrac{2}{x^2} - 1 + \dfrac{x^2}{3} \right) \left( 1 + \dfrac{x^2}{3} \right) \]

Expanding and simplifying:

\[ \approx \dfrac{2}{x^2} - \dfrac{1}{3} + \dfrac{x^4}{9} \]

Subtract \( \dfrac{2}{x^2} \):

\[ \left( \dfrac{2}{x^2} - \dfrac{1}{3} + \dfrac{x^4}{9} \right) - \dfrac{2}{x^2} = -\dfrac{1}{3} + \dfrac{x^4}{9} \]

Take the limit as \( x \to 0 \):

\[ \lim_{x \to 0} \left( -\dfrac{1}{3} + \dfrac{x^4}{9} \right) = -\dfrac{1}{3} \]

Thus, the limit is:

\[ \boxed{ \lim_{x \to 0} \left(\dfrac{1}{\sin^2 x} + \dfrac{1}{\tan^2 x} - \dfrac{2}{x^2}\right) = -\dfrac{1}{3} } \]

To find the limit \( \lim_{x \to 0} \dfrac{\pi \sin x - \sin(\pi x)}{x(\cos x - \cos(\pi x))} \), we use Taylor series expansions.

Expand the numerator using Taylor series:

\[ \pi \sin x \approx \pi\left(x - \dfrac{x^3}{6}\right), \quad \sin(\pi x) \approx \pi x - \dfrac{(\pi x)^3}{6} \]

Subtracting these:

\[ \pi \sin x - \sin(\pi x) \approx \dfrac{\pi(\pi^2 - 1)x^3}{6} \]

Expand the expressions between parentheses in the denominator using Taylor series:

\[ \cos x - \cos(\pi x) \approx \left(1 - \dfrac{x^2}{2}\right) - \left(1 - \dfrac{(\pi x)^2}{2}\right) = \dfrac{(\pi^2 - 1)x^2}{2} \]

Multiply by \( x \) to obtain the whole denominator:

\[ x(\cos x - \cos(\pi x)) \approx \dfrac{(\pi^2 - 1)x^3}{2} \]

Form the limit ratio:

\[ \dfrac{\dfrac{\pi(\pi^2 - 1)x^3}{6}}{\dfrac{(\pi^2 - 1)x^3}{2}} = \dfrac{\pi}{3} \]

Thus, the limit is:

\[ \boxed{ \lim_{x \to 0} \dfrac{\pi \sin x - \sin(\pi x)}{x(\cos x - \cos(\pi x))} = \dfrac{\pi}{3} } \]

Note: This limit could also be solved using L'Hôpital's rule by differentiating three times the numerator and denominator.

We shall find the limit \( \lim_{x \to 0} \dfrac{x + \ln(\sqrt{x^2 + 1} - x)}{x^3} \) using Taylor series expansions.

The binomial expansion for rational power is given by:

\[ (1 + \epsilon)^n = 1 + n\epsilon + \dfrac{n(n-1)}{2!} \epsilon^2 + \dfrac{n(n-1)(n-2)}{3!} \epsilon^3 + \cdots \]

Expand \( \sqrt{x^2 + 1} \) using binomial expansion:

\[ \sqrt{x^2 + 1} = (1 + x^2)^{1/2} = 1 + \dfrac{x^2}{2} - \dfrac{x^4}{8} + \cdots \]

Subtract \( x \) and take the natural logarithm:

\[ \ln(\sqrt{x^2 + 1} - x) = \ln\left(1 - x + \dfrac{x^2}{2} - \cdots\right) \]

Expand the logarithm using the Taylor series:

\[ \ln\left(1 - x + \dfrac{x^2}{2} - \cdots\right) \approx -x + \dfrac{x^3}{6} + \cdots \]

Combine with the \( x \) term in the numerator:

\[ x + \ln(\sqrt{x^2 + 1} - x) \approx x + \left(-x + \dfrac{x^3}{6}\right) = \dfrac{x^3}{6} \]

Divide the simplified numerator by the denominator \( x^3 \):

\[ \dfrac{\dfrac{x^3}{6}}{x^3} = \dfrac{1}{6} \]

Thus, the limit is:

\[ \boxed{ \lim_{x \to 0} \dfrac{x + \ln(\sqrt{x^2 + 1} - x)}{x^3} = \dfrac{1}{6} } \]

Note: This limit could also be solved using L'Hôpital's rule by differentiating three times the numerator and denominator.

To find the limit \( \displaystyle \lim_{x \to 0} \left( \dfrac{\sin x}{x} \right)^{\dfrac{1}{x^2}} \), we proceed as follows:

Let \( L \) be the limit to find:

\[ L = \lim_{x \to 0} \left( \dfrac{\sin x}{x} \right)^{\dfrac{1}{x^2}} \]

Take the natural logarithm of both sides to simplify the exponent:

\[ \ln L = \lim_{x \to 0} \ln \left( \dfrac{\sin x}{x}\right)^{\dfrac{1}{x^2}} \]

Simplify the right side:

\[ \ln L = \lim_{x \to 0} \dfrac{1}{x^2} \ln \left( \dfrac{\sin x}{x}\right) \]

Expand \( \sin x \) using Taylor series around \( x = 0 \):

\[ \sin x = x - \dfrac{x^3}{6} + \dfrac{x^5}{120} - \cdots \implies \dfrac{\sin x}{x} = 1 - \dfrac{x^2}{6} + \dfrac{x^4}{120} - \cdots \]

Approximate the logarithm using \( \ln(1 + y) \approx y - \dfrac{y^2}{2} + \cdots \) for small \( y \):

\[ \ln \left( \dfrac{\sin x}{x} \right) \approx \ln \left( 1 - \dfrac{x^2}{6} \right) \approx -\dfrac{x^2}{6} - \dfrac{x^4}{180} - \cdots \]

Substitute the approximation into the limit:

\[ \ln L = \lim_{x \to 0} \dfrac{-\dfrac{x^2}{6} - \dfrac{x^4}{180} - \cdots}{x^2} = \lim_{x \to 0} \left( -\dfrac{1}{6} - \dfrac{x^2}{180} - \cdots \right) = -\dfrac{1}{6} \]

Exponentiate the result to solve for \( L \):

\[ L = e^{-\dfrac{1}{6}} \]

And finally:

\[ \boxed{ L = \lim_{x \to 0} \left( \dfrac{\sin x}{x} \right)^{\dfrac{1}{x^2}} = e^{-\dfrac{1}{6}} } \]

To find the limit \( \displaystyle \lim_{x \to 0} \dfrac{\cos(\sin x) - \cos x}{x^4} \), we use Taylor series expansions for \( \sin x \) and \( \cos x \).

Taylor series expansions:

\[ \sin x = x - \dfrac{x^3}{6} + \cdots \] \[ \cos x = 1 - \dfrac{x^2}{2} + \dfrac{x^4}{24} - \cdots \]

Expand \( \cos(\sin x) \):

Substitute \( \sin x \) into the Taylor series for \( \cos(\theta) \) where \( \theta = \sin x \):

\[ \cos(\sin x) = \cos\left(x - \dfrac{x^3}{6} + \cdots\right) \]

Expand \( \cos(\theta) \) using \( \theta = x - \dfrac{x^3}{6} \):

\[ \cos(\theta) = 1 - \dfrac{\theta^2}{2} + \dfrac{\theta^4}{24} - \cdots \]

Calculate \( \theta^2 \) and \( \theta^4 \) up to \( x^4 \) terms:

\[ \theta^2 = \left(x - \dfrac{x^3}{6}\right)^2 = x^2 - \dfrac{x^4}{3} + \cdots \] \[ \theta^4 = \left(x - \dfrac{x^3}{6}\right)^4 = x^4 + \cdots \]

Substitute these into the expansion of \( \cos(\theta) \):

\[ \cos(\sin x) \approx 1 - \dfrac{x^2}{2} + \dfrac{5x^4}{24} + \cdots \]

Subtract \( \cos x \) from \( \cos(\sin x) \):

\[ \cos x \approx 1 - \dfrac{x^2}{2} + \dfrac{x^4}{24} + \cdots \]

Subtract the expansions:

\[ \cos(\sin x) - \cos x \approx \left(1 - \dfrac{x^2}{2} + \dfrac{5x^4}{24}\right) - \left(1 - \dfrac{x^2}{2} + \dfrac{x^4}{24}\right) = \dfrac{4x^4}{24} = \dfrac{x^4}{6} \]

Divide by \( x^4 \) and take the limit:

\[ \lim_{x \to 0} \dfrac{\dfrac{x^4}{6}}{x^4} = \dfrac{1}{6} \]

Thus, the limit is:

\[ \boxed{ \lim_{x \to 0} \dfrac{\cos(\sin x) - \cos x}{x^4} = \dfrac{1}{6} } \]