A limit problem? A limit question?

x->0

tanx = x +o(x)

ln[ 1+ (2/3)tanx ]=(2/3)x +o(x)

1+ (2/3)tanx ] =(2/3)x^2 +o(x^2)

e^【 =e^[ 2/3)x^2 +o(x^2)] = 1+ (2/3)x^2 +o(x^2)

e^【 1 =(2/3)x^2 +o(x^2)

1+(2/3)tanx]^x -1 =e^【 1 =(2/3)x^2 +o(x^2)

lim(x->0) [3+2tanx)^x -3^x ] / [ 3(sinx)^2 + x^3. cos(1/x) ]

lim(x->0) 3^x. / [ 3(sinx)^2 + x^3. cos(1/x) ]

lim(x->0) / [ 3(sinx)^2 + x^3. cos(1/x) ]

lim(x->0) (2/3)x^2/ [ 3(sinx)^2 + x^3. cos(1/x) ]

The numerator denominator is divided by x 2 at the same time

lim(x->0) (2/3)/ [ 3(sinx/x)^2 + x. cos(1/x) ]

You can't take a part of the addition of the numerator and replace it with the equivalent infinitesimal first, if it's a factor, it's okay, it's not an addition, this is a common mistake to find the limit, and you must figure it out. This question is best sought with Lobida.

<> this question is more convenient and easier to understand by using the power series of sinx.

From the question condition, it can be seen that the left = lim(x 0)(sin2x x +b x)+a=2 3.

Whereas, lim(x 0)(sin2x x +b x) = lim(x 0)(sin2x+bx) x. It belongs to the "0 0" type, using the rule of Lopida, lim(x 0)(sin2x x +b x) = lim(x 0)(2cos2x+b) (3x).

In this case, when x 0 and the denominator is 0, there must be 2cos2x+b 0. Again, 2cos2x+b is a continuous function, b=-2. ∴a=2/3。

f(x) (0->(sinx) 2) ln(1+t) dt e (2x 2)-2e (x 2) +1] sell; x ≠ 0 loose matter

a Brewing liquid; x=0

x->0

Denominator. e^(2x^2) =1 +2x^2 +(1/2)(2x^2)^2 +o(x^4)=1 +2x^2 +2x^4 +o(x^4)

2e (x 2) = 2 +2x 2 + x 4 + o(x 4) e (2x 2)-2e (x 2) +1 = x 4 + o(x 4) molecules. sinx)^2 = x^2 +o(x^2)(0->(sinx)^2) ln(1+t) dt∫(0->(sinx)^2) t dt

0->x^2) t dt

1/2)x^4

lim(x->0) ∫0->(sinx)^2) ln(1+t) dt / e^(2x^2)-2e^(x^2) +1]

lim(x->0) (1/2)x^4 / x^4a=1/2<>

Flexible use of the law of Lopida, stupid slag equivalence band quietly infinitesimal difference cavity substitution.

Use Taylor's formula to spread the pure jerk method at x=0 points.

sinx=x-1/3!*x^3+o(x^4)f(x)=f(0)+f'(0)*x+f''(0)/2!*x^2+o(x^3)

Then (sinx+xf(x)) x 3

x-1/3!*x^3+o(x^4)+f(0)*x+f'(0)*x^2+f''(0)/2!*x^3+o(x^4))/x^3

1+f(0))*x+f'(0)*x^2+(f''(0)/2!-1/3!)*x^3+o(x^4))/x^3

If the limit holds, and is equal to 1 2

Then it can only be.

1+f(0)=0

f'(0)=0

f''(0)/2!-1/3!=1/2

Hence f(0)=-1

f'(0)=0

f''(0)=(1/2+1/3!)*2!=1+1 3=4 3If you are satisfied, please choose the one [Satisfied]. Thank you ......If there is still something unclear, Hungry Bridge can ask [chase] for the age of pants.

In addition, after selecting [Satisfied] and other treatments, you can return the wealth value of the mortgage.

x 2 is derived from the equivalent infinitesimal substitution.

e^x-1)~x tanx~x

sin 2x x 2+(cosx-1) x 2 is still the result of an equivalent infinitesimal substitution.

sinx/x->1 cosx-1~-x^2/2 cosx-1/x^2->-1/2

So add it up to 1-1 2

If you don't understand, you can hi me.

Also note the references below.

Answer: Let y=(x+e x) (1 x), take the natural logarithm, have: lny=ln(x+e x) x

When calculating x tends to positive infinity, use Robida's rule to calculate the limit of lny (a denotes the sign that the limit approaches positive infinity when x approaches positive infinity).

a(lny)=a[(1+e^x)/(x+e^x)]/1=a[e^x/(1+e^x)]=a[e^x/e^x]=1

So: original limit = e

Omit the limit symbol. The original form can be written as.

1+x+e^x-1)^

Since the limit of [x+e x-1] x is 2, the original limit is e 2

X is ignored when it tends to be positive infinity.

e^x)^(1/x)=e

Click to enlarge and then click to enlarge again.

There is something wrong with this problem itself: the limit is 1, regardless of the value of c