Guide :

# How do you solve the limit ((1/(x+1))-1)/x as x approaches 0

How do you solve the limit?

## Research, Knowledge and Information :

### How do you solve the following limit (e^x-1)/x as x ...

It is a remarkable limit, but, if you want to demonstrate it, you have to know the fundamental limit: lim_(xrarroo)(1+1/x)^x=e (number of Neper), and also this limit ...

### What is the limit as x approaches 0 of 1/x^2? | Socratic

The exception to this rule is when taking the limit as x approaches c leads me to the ... What is the limit as x approaches 0 of #1/x ... How do you solve the ...

### Limits as x Approaches a Constant - UC Davis Mathematics

LIMITS OF FUNCTIONS AS X APPROACHES A CONSTANT ... many students INCORRECTLY conclude that is equal to 1 or 0 , or that the limit does not exist or is or .

### Lim x^{1/x}, x-> infty | Physics Forums - The Fusion of ...

So what is the limit of x^{1/x} as x approaches infinity? ... (as x-> infty), and limit of (1+x)^{1/x} = e (as x-> 0), ... One way to solve it is by observing that; x ...

### What is the limit as x approaches 1 of (2x^2 - x - 1) /(x-1 ...

The limit as x approaches 1 of this equation is 3. ... How do I solve [math](1+x) ... What is the limit as x approaches 0 of (x^2) /1-cosx? What is f(x) = ...

### Calculus Made Easy: How To Solve Calculus Limit Problems

How to Solve Calculus Limit Problems. ... The following limit does not exist as x approaches 2. ... To solve a limit that has the form 0/0, ...

### What is: [math]\displaystyle\lim_{x\to0} \frac {1}{x}[/math ...

... the limit as x approaches 0 for 1/x is undefined. You can test ... How do I solve the following limit: ... How do I find the limit [math]\displaystyle\lim_{x ...

### Limits to Infinity - Math Is Fun

Limits to Infinity. You should read Limits (An Introduction) first. ... The limit of 1 x as x approaches Infinity is 0. And write it like this: In other words:

### Limits - Evaluating

Limits (Evaluating) You should read Limits ... x (x 2 − 1) (x − 1) 0.5 : 1.50000: 0.9 : 1.90000: ... The limit of (x 2 −1) (x−1) as x approaches 1 is 2.

## Suggested Questions And Answer :

### how to solve this problem

It's a limit, so we're not concerned with what happens at x=1.  We're concerned with what happens as we approach x=1. The part of f(x) that talks about approaching x=1 is x^2 + 2.   If you plug x=1 into that you get (1)^2 + 2 = 3, so the limit of f(x) as x approached 1 is 3.

### can you solve this limit?

Let x=a+d where d is a very small positive or negative number much smaller than x. Put x=a+d into the limit expression: x^2-a^2=(x-a)(x+a)=d(2a+d)=2ad, if we ignore d^2 as being too small. sin(3x-3a)=sin(3d)=3d when d is very small. So the limit expression is 2ad/3d=2a/3. This is the result of applying the limit as d approaches zero, so x approaches a. If the limit is 4 then 2a/3=4 and a=3/2*4=6. So the limit is 6.

### How do you solve the limit ((1/(x+1))-1)/x as x approaches 0

{[ 1/(x+1)] -1}/x as x--> 0 [....] bekum [1-1*(x+1)]/(x+1)=-x/(x+1) divide this bi x, yu get -1/(x+1) as x-->0, yu get -1/1=-1

### how to graph a limit as x approaches 6 to the left when f(x)=3

This question is incomplete.

### finding average velocity

s=3t^2, (s'=velocity=6t). To work out the average velocity algebraically, we calculate s for t and t+h. s=3t^2 and 3(t+h)^2. The difference is 3(t+h-t)(t+h+t)=3h(2t+h)=6ht neglecting h^2 when h is very small. So the difference in distance is 6ht+3h^2 and the average velocity is (distance)/(time)=6t+3h. a i) Velocity at t=5 is 6*5=30, so 3h=0.3 making the average velocity=30.3. a ii) The average velocity is 30+3*0.01=30.03. a iii) The average velocity is 30.003. b. The instantaneous velocity at t=5 is 30 because h approaches zero. c. The limit definition is (6ht+3h^2)/h=6t+3h which becomes 6t as h approaches the limit of zero. So the instantaneous velocity at t=5 is 30.

### what is (a)?

Let x=a+d where d is a very small positive value compared to a. We can write x^2-a^2 as (a+d-a)(a+d+a)=d(2a+d)=2ad, if we ignore d^2 as insignificant. sin(3x-3a)=sin3(a+d-a)=sin3d. When d is small sin3d is approximately 3d. So we have 2ad/3d=2a/3. As d approaches 0, x approaches a, therefore the limit is 2a/3. Note that this is the same limit if d is negative, because d cancels out. I think this word explanation is sufficient not to require a picture or expansion in MS Word.

### how to solve limits of trigonometric function?

sine (1/x) at x gotu 0....infinity sine(x)=x -(1/3)x^3 +(1/5!)x^5+... sine(1/x)=(1/x)-+... as x gotu 0, 1/x gro without bound

### Find volume of resulting solid using any method

The picture below illustrates the problem. From this picture we can see that the tip of the sideways parabola is to be rotated around the axis y=5 that it is lying upon. When y=6, x=0 (vertex) and the part of the parabola we need is therefore between x=0 and 1. Since x=(y-6)^2, then y=6±√x. This gives us two circles for each x value. The larger circle has a radius of 6+√x-5=1+√x and the smaller circle has a radius of 1-√x. The difference between the areas of these two circles gives the area of a "washer": π((1+√x)^2-(1-√x)^2)=π(1+√x-1+√x)(1+√x+1-√x)=4π√x. The volume of the washer is 4π√xdx for an infinitesimally thin washer, thickness dx. If we integrate 4π∫(√xdx)=4π[(2/3)x^(3/2)] for 0≤x≤1, we get 8π/3=8.3776 cu units = volume of revolution. ALTERNATIVE APPROACH x=(y-6)^2 so √x=y-6 and dx=2(y-6)dy. The volume of the washer is 4π(y-6)dx=8π∫((y-6)^2dy) = 8π∫((y^2-12y+36)dy). The limits are found through solving x=1=(y-6)^2 making the intersections (1,5) and (1,7). But when x=0, y=6, so the limits are 6≤y≤7 not 5≤y≤7. Integrating: 8π[y^3/3-6y^2+36y] for 6≤y≤7=8π((7^3-6^3)/3-6(7^2-6^2)+36(7-6))=8π(127/3-78+36)=8π/3. The same result applies if we take the limits 5≤y≤6=8π((6^3-5^3)/3-6(6^2-5^2)+36(6-5))=8π(91/3-66+36)=8π/3. Had we used 5≤y≤7 we would have included the same volume twice, so the result would have been 16π/3.

### limit(1/2^1/2)^n ngoes to infinity

(1/2)^(1/2)=sqrt(1/2)=sqrt(2)/2<1 (approx 0.71). Because this is a fraction, as n gets larger the fraction gets smaller. When n=2 it's 1/2; when n=4, it's 1/4. As n approaches infinity the expression approaches zero.

### what is the limit as x approaches infinity of (x^3)*(sin(2pi/x))? and how to do it please

This shouls answer your question: http://www.wolframalpha.com/input/?i=lim%28x-%3Einfinity%29+%28x%5E3%29*sin%282pi+%2F+x%29   Click on the "Show Steps" button on the right hand side.