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Problem 41 Hard Difficulty

Find the limit. Use l'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. If l'Hospital's Rule doesn't apply, explain why.

$ \displaystyle \lim_{x\to 0} \frac{\cos x - 1 + \frac{1}{2}x^2}{x^4} $

Answer

This limit has the form $\frac{0}{0} . \lim _{x \rightarrow 0} \frac{\cos x-1+\frac{1}{2} x^{2}}{x^{4}} \stackrel{\underline{H}}{=} \lim _{x \rightarrow 0} \frac{-\sin x+x}{4 x^{3}} \underline{\underline{H}} \lim _{x \rightarrow 0} \frac{-\cos x+1}{12 x^{2}} \underline{\underline{h}} \lim _{x \rightarrow 0} \frac{\sin x}{24 x} \underline{\underline{H}} \lim _{x \rightarrow 0} \frac{\cos x}{24}=\frac{1}{24}$

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Video Transcript

We know that if we plug in, we end up with zero over zero, which implies that we should be using the hospital's rule so we can apply it on the end of it. The function. Negative sign expose acts over four x cubed. However, we still get zero over zero gonna plow. Hoppy tells you again. You still got zero over zero. Do it 1/3 time. The one is a constant effort, cancels. Okay, let's try it again. Okay. Now we have something we can work with. Co Sign of X gives us 1% of zero over 24. This is an answer. One over 24.