Solve the function limit problem, the function limit problem, how to find this

Solution: x tends to , lim(1 x + 2 (1 x)) x, take t=1 x, and the original formula becomes.

t tends to 0, lim(t + 2 t) (1 t) lim e ln[(t + 2 t) (1 t)] lim e [ln(t + 2 t) t], e [lim ln(t + 2 t) t].

Since ln(t+2 t) and t both tend to 0, using Robida's rule, lim ln(t + 2 t) t=lim[ln(t + 2 t)].'=lim(1+2 tln2) (t+2 tln2)=1+ln2, so the original formula =e (1+ln2)=2e

My idea is to think of it as a function, 1 x+2 1 x>1, the outer function is increasing, the inner function is decreasing, the total is decreasing, and when x is to positive infinity, the limit is 1

When x tends to infinity, 1 x is an infinitesimal amount compared to 2 (1 x), so.

x tends to infinity lim(1 x+2 (1 x)) x = lim(2 (1 x)) x = 2

It is enough to use the equivalent infinitesimal to do so.

Remember f(x)=(x+2 x) (1 x), and find the limit of f(x) when x 0.

At x 0, lnf(x)=ln(x+2 x) x x+2 x-1) x = 1+[exp(xln2)-1] x 1+xln2 x = 1+ln2

Therefore, when x 0, f(x) exp(1+ln2)=2e, is the conclusion!

Let y kx, simplify the original formula to get k (1 k), which shows that at the origin, it tends to the origin in different directions, and the limit obtained by the land is different from the jujube, because the trace does not exist in this original limit.

<> upper and lower limits are not equal, and the limits do not exist.

Let y kx, the original simplification of k (1 k), which means that at the origin, the origin is approached in different directions, and the resulting limit is different, so the original limit does not exist.

1 All (1) up and down the same as dividing x 2

lim(1+1/x-3/x^2)/3(1-1/x)^2=(1+0-0)/3(1-0)^2

2) Divide up and down by x 3

lim(2/x)/(3-1/x^2+9/x^3)=0/(3-0+0)

0 (3) divides x by the same up and down

lim(x-4-7 x) (1-8 x)=(infinity-4-0) 1

Infinity (4) divides by (2x) 50

lim(1-1/2x)^20(1/(2x)^29+6/(2x)^30)/(1-3/2x)^50

0(5) The numerator is physicochemical, and the numerator and denominator are multiplied by the root number (x 2-1) + x = lim [x(root number (x 2-1)-x)(root number (x 2-1) +x)] (root number (x 2-1) + x).

lim x(-1) (root number (x 2-1) + x) is divided by x

lim (-1) [root number (1-1 x 2)+1] = -1 (1+1).

6) Molecules are physicochemical.

The numerator and denominator are multiplied by the root number (x 4-1) + x 2

lim (root number (x 4-1) - x 2) (root number (x 4-1) + x 2) (root number (x 4-1) + x 2).

lim (-1) (root number (x 4-1) + x 2) = -1 (infinity + infinity) = 0

The first numerator and denominator are divided by x to the second power, and then the limits of 1 x and 1 x2 are all zero, and the result is 1 3

Other similar problems need to be thought about first, and the numerator is the same differentiation, for example, the numerator and denominator of the fifth question are multiplied by x2-1+x at the same time

In this question, we should pay attention to the rank of the infinity term in the numerator denominator, and the exponent tends to be dominant in the infinity polynomial. The specific process is as follows: