Find Limits Using Series
Question 1
Find the limit \[ \lim_{x \to 0}\frac{1-\frac12 x^2 - \cos\left(\frac{x}{1-x^2}\right)}{x^4} \]
Solution
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:
\[
\frac{x}{1 - x^2} = x \left(1 + x^2 + x^4 + \cdots \right) = x + x^3 + x^5 + \cdots
\]
For small \(x\), we approximate this as \(\theta = x + x^3\).
Expand \(\cos(\theta)\) using the Taylor series:
\[
\cos(\theta) = \cos(x + x^3) \approx 1 - \frac{\theta^2}{2} + \frac{\theta^4}{24} - \cdots
\]
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:
\[
\cos(x + x^3) \approx 1 - \frac{x^2 + 2x^4}{2} + \frac{x^4}{24} = 1 - \frac{x^2}{2} - x^4 + \frac{x^4}{24}
\]
Simplify the terms:
\[
\cos(x + x^3) \approx 1 - \frac{x^2}{2} - \frac{23x^4}{24}
\]
Substitute this back into the numerator of the original expression:
\[
1 - \frac{1}{2}x^2 - \cos\left(\frac{x}{1 - x^2}\right) \approx 1 - \frac{1}{2}x^2 - \left(1 - \frac{x^2}{2} - \frac{23x^4}{24}\right)
\]
and simplify it
\[
1 - \frac{1}{2}x^2 - 1 + \frac{1}{2}x^2 + \frac{23x^4}{24} = \frac{23x^4}{24}
\]
Divide the simplified numerator by the denominator \(x^4\) and simplify:
\[
\frac{\frac{23x^4}{24}}{x^4} = \frac{23}{24}
\]
Thus, the limit is \[ \boxed{ \lim_{x \to 0}\frac{1-\frac12 x^2 - \cos\left(\frac{x}{1-x^2}\right)}{x^4} = \dfrac{23}{24}}\].
Question 2
Find the limit
\[ \lim_{x\to 0} \left(\frac 1{\sin^2 x} + \frac 1{\tan^2x} -\frac 2{x^2} \right) \]
Solution
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 :
\[
\frac{1}{\tan^2 x} = \frac{\cos^2 x}{\sin^2 x}
\]
Thus, the expression becomes:
\[
\frac{1}{\sin^2 x} + \frac{\cos^2 x}{\sin^2 x} - \frac{2}{x^2} = \frac{1 + \cos^2 x}{\sin^2 x} - \frac{2}{x^2}
\]
Expand \(\cos x\) and \(\sin x\) using Taylor series :
\[
\cos x \approx 1 - \frac{x^2}{2} + \frac{x^4}{24}
\]
\[
\sin x \approx x - \frac{x^3}{6}
\]
Squaring the above expansions:
\[
\cos^2 x \approx \left(1 - \frac{x^2}{2} + \frac{x^4}{24}\right)^2 \approx 1 - x^2 + \frac{x^4}{3}
\]
\[
\sin^2 x \approx \left(x - \frac{x^3}{6}\right)^2 \approx x^2 - \frac{x^4}{3}
\]
Substitute these expansions into the expression:
\[
1 + \cos^2 x \approx 2 - x^2 + \frac{x^4}{3}
\]
\[
\frac{1 + \cos^2 x}{\sin^2 x} \approx \frac{2 - x^2 + \frac{x^4}{3}}{x^2 - \frac{x^4}{3}} = \frac{2 - x^2 + \frac{x^4}{3}}{x^2 \left(1 - \frac{x^2}{3}\right)}
\]
Simplifying:
\[
\frac{2 - x^2 + \frac{x^4}{3}}{x^2 \left(1 - \frac{x^2}{3}\right)} \approx \left( \frac{2}{x^2} - 1 + \frac{x^2}{3} \right) \left( 1 + \frac{x^2}{3} \right)
\]
Expanding and simplifying:
\[
\approx \frac{2}{x^2} - \frac{1}{3} + \frac{x^4}{9}
\]
Subtract \(\frac{2}{x^2}\) from the a:
\[
\left( \frac{2}{x^2} - \frac{1}{3} + \frac{x^4}{9} \right) - \frac{2}{x^2} = -\frac{1}{3} + \frac{x^4}{9}
\]
Take the limit as \(x \to 0\):
\[
\lim_{x \to 0} \left( -\frac{1}{3} + \frac{x^4}{9} \right) = -\frac{1}{3}
\]
Thus, the limit is \[ \boxed{\lim_{x\to 0} \left(\frac 1{\sin^2 x} + \frac 1{\tan^2x} -\frac 2{x^2} \right) = -\frac{1}{3}} \].
Question 3
Find the limit
\[ \lim_{x \to 0} \frac{\pi \sin x - \sin(\pi x)}{x(\cos x - \cos(\pi x))} \]
Solution
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 :
\[
\pi \sin x \approx \pi\left(x - \frac{x^3}{6}\right), \quad \sin(\pi x) \approx \pi x - \frac{(\pi x)^3}{6}
\]
Subtracting these:
\[
\pi \sin x - \sin(\pi x) \approx \frac{\pi(\pi^2 - 1)x^3}{6}
\]
Expand the expressions between parentheses in the denominator using Taylor series :
\[
\cos x - \cos(\pi x) \approx \left(1 - \frac{x^2}{2}\right) - \left(1 - \frac{(\pi x)^2}{2}\right) = \frac{(\pi^2 - 1)x^2}{2}
\]
Multiply by \(x\) to obtain the whole denominator:
\[
x(\cos x - \cos(\pi x)) \approx \frac{(\pi^2 - 1)x^3}{2}
\]
Form the limit :
\[
\frac{\dfrac{\pi(\pi^2 - 1)x^3}{6}}{\dfrac{(\pi^2 - 1)x^3}{2}} = \dfrac{\pi}{3}
\]
Thus, the limit is \[ \boxed{\lim_{x \to 0} \frac{\pi \sin x - \sin(\pi x)}{x(\cos x - \cos(\pi x))} = \dfrac{\pi}{3}} \]
Note
The limit in question 3 could also be solved using L'Hopital theorem by differentiating three times the numerator and denominator.
Question 4
Find the limit
\[ \lim_{x \to 0} \frac{x + \ln(\sqrt{x^2 + 1} - x)}{x^3} \]
Solution
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: \[ (1+\epsilon)^n = 1+n \epsilon + \dfrac{n(n-1)}{2!} \epsilon^2 + \dfrac{n(n-1)(n-1)}{3!} \epsilon^3 +... \]
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:
\[
\ln(\sqrt{x^2 + 1} - x) = \ln\left(1 - x + \frac{x^2}{2} - \cdots\right)
\]
Expand the logarithm using the Taylor series:
\[
\ln(1 - x + \frac{x^2}{2} - \cdots) \approx -x + \frac{x^3}{6} + \cdots
\]
Combine with the \(x\) term in the numerator:
\[
x + \ln(\sqrt{x^2 + 1} - x) \approx x + (-x + \frac{x^3}{6}) = \frac{x^3}{6}
\]
Divide the simplified numerator by the denominator \(x^3\):
\[
\dfrac{\frac{x^3}{6}}{x^3} = \dfrac{1}{6}
\]
Thus, the limit is \[ \boxed{ \lim_{x \to 0} \frac{x + \ln(\sqrt{x^2 + 1} - x)}{x^3}= \dfrac{1}{6}}\]
Note
The limit in question 4 could also be solved using L'Hopital theorem by differentiating three times the numerator and denominator.
Question 5
Find the limit
\[ \lim_{x \to 0} \left( \frac{\sin x}{x} \right)^{\frac{1}{x^2}} \]
Solution
To find the limit \(\displaystyle \lim_{x \to 0} \left( \frac{\sin x}{x} \right)^{\frac{1}{x^2}}\), we proceed as follows:
Let \( L \) be the limit to find:
\[
L = \lim_{x \to 0} \left( \frac{\sin x}{x} \right)^{\frac{1}{x^2}}
\]
Take the natural logarithm of both sides to simplify the exponent:
\[
\ln L = \lim_{x \to 0} \ln \left( \frac{\sin x}{x}\right)^{\frac{1}{x^2}}
\]
Simplify the right side:
\[
\ln L = \lim_{x \to 0} {\frac{1}{x^2}} \ln \left( \frac{\sin x}{x}\right)
\]
Expand \(\sin x\) using Taylor series around \(x = 0\):
\[
\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots \implies \frac{\sin x}{x} = 1 - \frac{x^2}{6} + \frac{x^4}{120} - \cdots
\]
Approximate the logarithm using \(\ln(1 + y) \approx y - \frac{y^2}{2} + \cdots\) for small \(y\):
\[
\ln \left( \frac{\sin x}{x} \right) \approx \ln \left( 1 - \frac{x^2}{6} \right) \approx -\frac{x^2}{6} - \frac{x^4}{180} - \cdots
\]
Substitute the approximation into the limit:
\[
\ln L = \lim_{x \to 0} \frac{-\frac{x^2}{6} - \frac{x^4}{180} - \cdots}{x^2} = \lim_{x \to 0} \left( -\frac{1}{6} - \frac{x^2}{180} - \cdots \right) = -\frac{1}{6}
\]
Exponentiate the result to solve for \(L\):
\[
L = e^{-\frac{1}{6}}
\]
and finally
\[
\boxed{ L = \lim_{x \to 0} \left( \frac{\sin x}{x} \right)^{\frac{1}{x^2}} = e^{-\frac{1}{6}} }
\]
Question 6
Find the limit
\[ \lim_{x \to 0} \frac{\cos(\sin x) - \cos x}{x^4} \]
Solution
To find the limit \(\displaystyle \lim_{x \to 0} \frac{\cos(\sin x) - \cos x}{x^4}\), we use Taylor series expansions for \(\sin x\) and \(\cos x\).
Taylor series expansions :
\(\sin x = x - \frac{x^3}{6} + \cdots\)
\(\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots\)
Expand \(\cos(\sin x)\) :
Substitute \(\sin x\) into the Taylor series for \(\cos \theta\) where \(\theta = \sin x\):
\[
\cos(\sin x) = \cos\left(x - \frac{x^3}{6} + \cdots\right)
\]
Expand \(\cos(\theta)\) using \(\theta = x - \frac{x^3}{6}\):
\[
\cos(\theta) = 1 - \frac{\theta^2}{2} + \frac{\theta^4}{24} - \cdots
\]
Calculate \(\theta^2\) and \(\theta^4\) up to \(x^4\) terms:
\[
\theta^2 = \left(x - \frac{x^3}{6}\right)^2 = x^2 - \frac{x^4}{3} + \cdots
\]
\[
\theta^4 = \left(x - \frac{x^3}{6}\right)^4 = x^4 + \cdots
\]
Substitute these into the expansion of \(\cos(\theta)\):
\[
\cos(\sin x) \approx 1 - \frac{x^2}{2} + \frac{5x^4}{24} + \cdots
\]
Subtract \(\cos x\) from \(\cos(\sin x)\) :
\(\cos x \approx 1 - \frac{x^2}{2} + \frac{x^4}{24} + \cdots\)
Subtract the expansions:
\[
\cos(\sin x) - \cos x \approx \left(1 - \frac{x^2}{2} + \frac{5x^4}{24}\right) - \left(1 - \frac{x^2}{2} + \frac{x^4}{24}\right) = \frac{4x^4}{24} = \frac{x^4}{6}
\]
Divide by \(x^4\) and take the limit :
\[
\lim_{x \to 0} \frac{\frac{x^4}{6}}{x^4} = \frac{1}{6}
\]
Thus, the limit is \[ \boxed{ \lim_{x \to 0} \frac{\cos(\sin x) - \cos x}{x^4} = \dfrac{1}{6}} \].
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