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Find the limit of LIM (e ^ 1 / x + 1) / (e ^ 1 / x-1) arctan1 / X when x tends to 0?
The strict answer is as follows:
Find the limit Lim [e ^ x-e ^ (- x)] / X → 0 I haven't learned the law of Robita... Can I use other methods? Or make it clear Can this problem turn it into the form of the important limit limsinx / x? We only learn here so far Taylor unfolded... I found that my IQ was not generally low Why doesn't our teacher tell us anything!
0 / 0 type, this problem is solved by using the Robida's law. It is relatively simple to obtain the derivative of the numerator and denominator respectively: Lim [e ^ x + e ^ (- x)] / 1 x → 0 = 1 + 1 = 2. Taylor expansion of e ^ X and e ^ - x respectively (analogy equivalent to infinitesimal) obtains that e ^ x = 1 + X + O (x), e ^ - x = 1-x + O (x) are reduced to 2x, so the above limit is lim2x / x = 2
Find the limit LIM (x tends to 0) (x-e ^ x) ^ (2 / x)
I think there is something wrong with the formula you give. Y = x is less than y = e ^ X in any case, so the power in parentheses is always less than zero. A negative power does not always exist. It's strange
Find the limit LIM (1 / x2-1 / (xtanx)) x → 0
lim (1/x^2 - 1/xtanx)
=LIM (xtanx - x ^ 2) / (x ^ 3 * TaNx) about a TaNx
=LIM (TaNx - x) / x ^ 3 find derivatives up and down (ropida's law)
= lim (secx * secx -1)/3x^2
= lim (tanx * tanx)/3x^2
= 1/3
Find the limit: LIM (1 / x2-cot2x) x → 0 What's wrong with this: LIM (sin2x-x2cos2x) / x2sin2x = LIM (x2-x2 (1-sin2x)) / X4 = LIM (x2-x2 + x4) / X4 = 1 Why not?
Guess from your problem solving process that the original problem may be X - > 0, Lim [1 / x ^ 2 - (Cotx) ^ 2]
The reason for the error is that the equivalent infinitesimal can only replace the factor. There is a problem with the replacement numerator above you. The replacement of the denominator is correct
lim[ 1/x^2 - (cotx)^2]
=Lim [1 / x ^ 2 - (CSCX) ^ 2] + 1 (use (CSCX) ^ 2 = (Cotx) ^ 2 + 1)
=Lim [(SiNx) ^ 2 - x ^ 2] / [x ^ 2 * (SiNx) ^ 2] + 1 (general score)
=Lim [(SiNx) ^ 2 - x ^ 2] / (x ^ 4) + 1 (denominator Equivalent Infinitesimal Replacement)
=Lim [2sinxcosx - 2x] / (4x ^ 3) + 1 (lobita's law)
=Lim [sin2x - 2x] / (4x ^ 3) + 1 (trigonometric formula sin2x = 2sinxcosx)
=Lim [2cos2x - 2] / (12x ^ 2) + 1 (lobida's law)
= lim(cos2x-1)/ (6x^2) +1
=LIM (- 2sin2x) / (12x) + 1 (lobitar's law, also equivalent to infinitesimal)
=lim(-4x)/(12x) +1
=2/3
SiNx squared divided by the square of SiNx. What is the limit when x approaches 0? There must be a process,
lim[x→0](sinx)^2/(sinx)^2
=lin[x→0][ (sinx)^2/x^2]/[(sinx)^2/x^2]
=1/1=1
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