Use series to evaluate the limit #\lim_(x\rarr0)(x^2/2-1-\cos(x))/x^4#?

It says #x^2#/#2# in my worksheet. Not sure if it's supposed to be formatted the way it did above.

Anyway... how do I go about doing this?

Answer 1

Using #lim_(x-> 0) (x^2/2 - 1 + cosx)/x^4#, we get a result of #1/24#.

Recall that #cosx = 1 - x^2/2 +x^4/(4!) + ...#
#L = lim_(x->0) (x^2/2 - 1 - (1 - x^2/2 + x^4/(4!) + ...))/x^4#
This clearly evaluates to nothing but #-oo#. However, if the question was #(x^2/2 - 1 + cosx)/x^4#, we would see that
#L = lim_(x->0) (x^2/2 - 1 + 1 - x^2/2 + x^4/(4!) + ...)/x^4#
#L = lim_(x->0) (x^4/(4!) - x^6/(6!) + x^8/(8!) + ....)/x^4#
#L = 1/(4!) + 0 + 0 + 0 + ...#
#L = 1/24#

Hopefully this helps!

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Answer 2

#-oo#

Recall the Maclaurin Series expansion for cosine,

#cosx=sum_(n=0)^oo(-1)^nx^(2n)/((2n)!)=1-x^2/2+x^4/(4!)-x^2/(6!)+...#

We can insert this series into our limit (that is, the first few terms of the series, followed by the ellipsis to indicate that it goes on forever):

#lim_(x->0)(x^2/2-1-(1-x^2/2+x^4/(4!)-x^2/(6!)+...))/x^4#

And distribute the negative sign, allowing us to combine the relevant terms:

#lim_(x->0)(x^2/2-1-1+x^2/2-x^4/(4!)+x^6/(6!)-...)/x^4#
#lim_(x->0)((cancel2x^2)/cancel2-2-x^4/(4!)+x^6/(6!)-...)/x^4#
#lim_(x->0)(x^2-2-x^4/(4!)+x^6/(6!)-...)/x^4#
Divide all terms by #x^4:#
#lim_(x->0)1/x^2-2/x^4-1/(4!)+x^2/(6!)-...#
We see that #x^2/(6!)# vanishes as #x->0,# and so would all terms that come after it, as all the following terms would have an #x# raised to increasingly higher powers in the numerator.

So, since all of the subsequent terms vanish, we won't include ellipsis anymore as we're no longer dealing with them.

Perform some simplification on the first two terms to take the limit:

#1/x^2-2/x^4=(x^4-2x^2)/x^6=(x^2(x^2-2))/x^6#
#=(x^2-2)/(x^4)#
Take the limit, and note that the constant #-1/(4!)# ends up having no impact due to the infinite nature of the limit:
#lim_(x->0)(x^2-2)/(x^4)-1/(4!)=(-2)/0-1/(4!)=-oo#
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Answer 3

To evaluate the limit (\lim_{x \to 0} \frac{x^2/2 - 1 - \cos(x)}{x^4}), we can use series expansion.

First, expand (\cos(x)) using its Maclaurin series: [ \cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \ldots ]

Now substitute this expansion into the original expression: [ \lim_{x \to 0} \frac{\frac{x^2}{2} - 1 - \left(1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \ldots \right)}{x^4} ]

Simplify the numerator: [ \lim_{x \to 0} \frac{\frac{x^2}{2} - 1 + \frac{x^2}{2!} - \frac{x^4}{4!} + \frac{x^6}{6!} - \ldots}{x^4} ]

Combine like terms: [ \lim_{x \to 0} \frac{\left(\frac{x^2}{2} + \frac{x^2}{2!}\right) - 1 - \frac{x^4}{4!} + \frac{x^6}{6!} - \ldots}{x^4} ]

Now simplify further and cancel out common terms: [ \lim_{x \to 0} \frac{\frac{x^2}{2} \left(1 + \frac{1}{1!}\right) - 1 - \frac{x^4}{4!} + \frac{x^6}{6!} - \ldots}{x^4} ]

The terms involving (x^4) and higher powers in the numerator will approach zero as (x) approaches zero. Therefore, we can ignore them in the limit calculation: [ \lim_{x \to 0} \frac{\frac{x^2}{2} \left(1 + \frac{1}{1!}\right) - 1}{x^4} = \lim_{x \to 0} \frac{\frac{3}{2}x^2 - 1}{x^4} ]

Now simplify and evaluate the limit: [ \lim_{x \to 0} \frac{\frac{3}{2}x^2 - 1}{x^4} = \lim_{x \to 0} \frac{\frac{3}{2} - \frac{1}{x^2}}{x^2} ]

As (x) approaches zero, the term (\frac{1}{x^2}) approaches infinity, making the fraction (\frac{3}{2} - \frac{1}{x^2}) approach (\frac{3}{2}). Therefore, the limit is: [ \lim_{x \to 0} \frac{\frac{3}{2} - \frac{1}{x^2}}{x^2} = \frac{3}{2} ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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