Calculation limit LIM (x → 1) x * 2-1 / X-1

Calculation limit LIM (x → 1) x * 2-1 / X-1

lim(x→1)x^2-1/x-1
=lim(x→1)x^2-1/x-1
lim(x→1)(x+1）
=2

Let f (x) = 1 / 3x & # 179; - X & # 178; + (1-A) x + 1, where a > 0 If a = 1, find the curve y=
The known function f (x) = 1 / 3x & # 179; - X & # 178; + (1-A) x + 1, where a > 0
First question If a = 1, find the tangent equation of curve y = f (x) at point (1, f (1));
It's better to take pictures

In this paper, f (x) = 1 / 3x3 + 1 / 2 (2-A) x2 + (1-A) x = > F (x) = 1 / 3x [x2 + 3 / 2 (2-A) x + 3 (1-A)] Let G = 1 / 3x; y = x2 + 3 / 2 (2-A) x + 3 (1-A) and then convert it into the standard formula, which is y = a (x + B / 2a) 2 - (b2-4ac) / 4A ------ (because there is a small case a in the title, the standard formula is all in big letters)

The problem that addition and subtraction can't be used in Equivalent Infinitesimal Substitution
Some people say that if the two formulas before and after replacement are infinitesimal equivalent, some people say that the two formulas of addition and subtraction require a limit (I think it's strange, for example, when x tends to 0, (TaNx SiNx) / X is not OK, SiNx and TaNx both have limits and can't be replaced). Can they be used, and what are the conditions

In the process of finding the limit of infinitesimal to infinitesimal, a factor in the numerator or denominator can be replaced by an equivalent infinitesimal. Generally, the equivalent infinitesimal can not be replaced by an equivalent infinitesimal in addition and subtraction. The condition of Equivalent Infinitesimal Replacement in addition and subtraction is that the limit in Lima / b exists and the limit is not equal to - 1, then the infinitesimals A and B in a + B can be replaced by their equivalent infinitesimal

How to find LIM (LN tan7x) / (LN tan2x) when x tends to 0 +,

=LIM (LN sin7x - ln cos7x) / (LN sin2x - ln cos2x) = LIM (LN sin7x - 0) / (LN sin2x - 0) = LIM (LN sin7x) / (LN sin2x)

How to find the limit of (LN tan7x) / (LN tan2x) when x tends to 0 +?

If x tends to zero, then Tan ~ X
And LNX tends to infinity
So the original formula = limln7x / ln2x
=lim(ln7+lnx)/(ln2+lnx)
Up and down divided by LNX
=lim(ln7/lnx+1)/(ln2/lnx+1)
=1

Lim ln (tan3x) / ln (tan2x) when x → 0 +

According to the law of Robita, the original formula is LIM (1 / tan3x · sec & # 178; 3x · 3) / (1 / tan2x · sec & # 178; 2x · 2) = LIM (3tan2x) / (2tan3x) · LIM (SEC & # 178; 3x / sec & # 178; 2x) = 3 / 2 · LIM (2x) / (3x) · 1 = 3 / 2 · 2 / 3 = 1

Seeking limit by law of lobida
LIM (x approaches 0 +) lntan7x / lntan3x
Using the law of lobita,

This is the ratio of infinity to infinity. According to the law of lobita, the original formula = (1 / tan7x * (tan7x) '/ (1 / tan3x * (tan3x)') = (1 / tan7x * 7 / (cos7x) ^ 2) / (1 / tan3x * 3 / (cos3x) ^ 2) = (7 / (sin7x * cos7x)) / (3 / (sin3x * cos3x) is divided by X to get the original formula = LIM (x approaches 0 +)

Find the limit of limx tending to 0 xcotx. Use the law of lobita

When x approaches 0, find the limit of (4 + x) - 2 divided by sin2x under the root sign, and find the specific solution
Find the concrete solution and application formula.

Lim X - > 0 [radical (x + 4) - 2] / sin2x
=Lim X - > 0 [radical (x + 4) - 2] / sin2x
=Lim X - > 0 X / [sin2x * (radical (x + 4) + 2)]
2x sin2x
=>The above formula = 1 / [2 * 4] = 1 / 8

What is the limit of x-sin2x / x + sin5x when x approaches 0?

X + sin5x when x approaches 0, the limit is 0 (just substitute it directly)
-Sin2x / X is of type 0 / 0. When x approaches 0, SiNx / x = 1
If x approaches 0, sin2x / 2x = 1,
So - sin2x / x = - 2
The limit of x-sin2x / x + sin5x when x approaches 0 is - 2