Vtyjkyuk

I want to evaluate the following limit:

$$lim_xrightarrow0frac(x-arctan(x))ln(1+2sin(x))(1+cosx)(e^x-1-x)^2$$

For example, we have $x-arctanx$. They are both $0$. This seems to be the so called "cancellation of terms". Therefore I apply the Taylor series:

$$arctanx=x-fracx^33+o(x^3)$$

Therefore

$$x-arctanxsim x-x +fracx^33+o(x^3)=fracx^33+o(x^3)$$

Somebody solved this problem textbook and developed this to the fifth term (not third). Now a few questions arise:

• Why do I need to develop this Taylor series to the fifth term (assuming I have to).


• Do I need to develop every function to the same term in my limit?

Any hints?

For the direct questions: One must, needs to, determine powers of the expansion beyond constant factors to determine the behavior of the expansions; Not every function has the same powers in an expansion. The result then becomes "at least get a few powers of $x$ of each function".
This becomes more familiar by the example in question. Consider each of the four functions first:
beginalign
x - tan^-1(x) &= fracx^33 - fracx^55 + fracx^77 + mathcalO(x^9) \
ln(1 + 2 , sin(x)) &= 2 x - 2 x^2 + frac7 x^33 - frac10 x^43 + mathcalO(x^5) \
1 + cos(x) &= 2 cos^2left(fracx2right) = 2 - fracx^22 + fracx^424 + mathcalO(x^6) \
(e^x - 1 - x)^2 &= fracx^44 + fracx^56 + frac5 x^672 + mathcalO(x^7).
endalign
With these expansions it is seen that the growth of powers of $x$ are not the same, but at least each expansion has powers of $x^4$ or greater.

Now, for the limit.
beginalign
F(x) &= frac(x - tan^-1(x) ) , ln(1 + 2 , sin(x))2 , cos^2left(fracx2right) , (e^x - 1 - x)^2 \
&= fracleft(fracx^33 - fracx^55 + fracx^77 + cdotsright) left( 2 x - 2 x^2 + frac7 x^33 - frac10 x^43 + cdotsright)left(2 - fracx^22 + fracx^424 + cdots right) left(fracx^44 + fracx^56 + frac5 x^672 + cdots right) \
&= fracfrac2 x^43 - frac2 x^53 + frac17 x^645 + cdotsfracx^42 + fracx^56 + fracx^672 + cdots \
&= fracfrac23 - frac2 x3 + frac17 x^245 + cdotsfrac12 + fracx6 + fracx^272 + cdots \
&= frac43 , left( 1 - frac4 x3 + frac49 x^236 + cdots right)
endalign
and
beginalign
lim_x to 0 F(x) &= lim_x to 0 frac43 , left( 1 - frac4 x3 + frac49 x^236 + mathcalO(x^3) right) \
&= frac43
endalign

You don't need to expand every function in your fraction to the same limit. For example, $1+cos xto 2$ as $xto 0$ so you don't need to anything at all here.

Thus, the only problem is now $$(e^x-1-x)^2 = left(fracx^22+o(x^2)right)^2 = x^4left(frac 14+o(x)right)$$
This means you just need to match this up in your numerator and it should be obvious from this point on.

1.i have used the expansion of e^x , as have used the standsr limits.
2. you expand till the terms, such that order if zero in numerator and denominator can be compared to one and other
i have missed 2 in denominator