376. lim (sqrt1+x-sqrt1-x)/x, x o0 377. lim (e^x-x-1)/x^2,...

376. lim (sqrt{1+x}-sqrt{1-x})/x, x o0 377. lim (e^x-x-1)/x^2, x o0 378. lim tan x/sqrt{x}, x o0 379. lim (x-1)/ln x, x o1 380. lim (x+1)^{1/x}, x o0 381. lim (sqrt{x}-sqrt[3]{x})/(x-1), x o1 382. lim x^{2x}, x o0^+ 383. lim x sin(1/x), x oinfty 384. lim (sin x-x)/x^2, x o0

Transcript

00:01 Hi there, in this question, there are two many limit questions we have.

00:06 So i'm going to just solve some of them.

00:08 So the first one, let me just write the first one, which is the limit when x goes to zero.

00:14 We have one plus x minus one minus x divided by x.

00:20 So the first thing that we have to do for the limit questions, we have to plug in the value where the x goes to.

00:27 So in this case, x goes to zero.

00:30 I'm going to just plug in 0 here, which is 1 minus 1 over 0, which is equal to 0 over 0 in determinant form.

00:36 So in this kind of indeterminate form, we have two options.

00:40 The first one is taking the derivative of the top and the bottom, which is the low pita rule.

00:46 And also we can just use the gathered of this undeterminate form by just taking the conjugate of the functions.

00:57 So first of all, i'm going to just multiply the top and the.

01:00 The bottom of the fraction by 1 plus x minus 1 minus x so multiply this by 1 plus x plus 1 minus x and divide the bottom which is x times radical 1 plus x plus radical 1 minus x so and on the top of the fraction we have two so difference of the squares which is 1 plus x minus 1 minus x divided by x times radical 1 plus x plus 1 minus x so if i simplify the top of the fraction i will have i just expand the parentheses and we got 2x on the top and on the bottom we have x times this is radical 1 plus x and plus radical 1 minus x that we have so these x are cancelled to each other so what is left i'm gonna just plug in 0 for the remaining thing which is 2 over this is 1 plus 1 that is equal to 2 over 2, which is equal to 1.

02:04 So this is the limit equation of this function here.

02:09 And what else? v22? so for the second.

02:11 So you can also solve the first question by using the lopital rule.

02:15 Maybe this is more practical way to solve in this way.

02:18 So we can just apply the next questions, the lopital rule, which is e to the power x minus 1 over x squared.

02:25 So if i plug in 0, again, the number is plugging 0 here, 1.

02:31 1 minus 1, which is 0 over 0.

02:33 Again, we have the indeterminate form...

Question

376. lim (sqrt{1+x}-sqrt{1-x})/x, x o0 377. lim (e^x-x-1)/x^2, x o0 378. lim tan x/sqrt{x}, x o0 379. lim (x-1)/ln x, x o1 380. lim (x+1)^{1/x}, x o0 381. lim (sqrt{x}-sqrt[3]{x})/(x-1), x o1 382. lim x^{2x}, x o0^+ 383. lim x sin(1/x), x oinfty 384. lim (sin x-x)/x^2, x o0

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