# Math Help - one sided limit help me..

1. ## one sided limit help me..

find one sided limit

1.)lim x->3+ X^2-4X+6 / square root x- 3

2.)lim t->0+ square root 4+t -2 / square root t

can someone teach me how to solve this kind of question thx

2. Hello. I'm new here, however, I think that I can help you. The answers to your questions are 1) DNE (infinity) and 2) 0.

Neither of these functions are defined on the other side of the limiting value, so you are forced to analyze the limits from only one side. This is fine. After all, one-sided limits are defined along the same criteria as two-sided limits. Without going into a formal epsilon-delta definition, a limit exists if there is convergence as you near the limiting value. That is, if a function grows excessively large, excessively small, or continually oscillates between values as you approach the limiting value, its limit at that limiting value does not exist.

The first function's limit is infinity because as you approach 3 the function grows without bound. The second function's limit is 0 because that is where the function converges as T approaches 0. The first conclusion was reached through simple direct substitution. The second function is a bit trickier because you need to remove a one-sided discontinuity through rationalization.

3. When you have a problem that has a numerator that doesn't equal zero over a denominator that equals zero, your limit will approach infinity.

So you can solve the first one numerically.

As x gets closer and closer to three from the right side, the denominator becomes infinitely large, thus as x approaches 3 from the right side, the fraction becomes infinite in size. The answer to #1 = infinity.

4. thank you

but i still dont understand,can u explain it using the calculation to solve the problem..

5. Originally Posted by ucen158
thank you

but i still dont understand,can u explain it using the calculation to solve the problem..
Sure. I haven't learned to use the math program yet, so bare with me. The first step in solving any limit problem is simple substitution. If you plug a limiting value into a function and get a value of some kind, then that's the limit. Period. This works with a bevy of functions including: polynomials, rationals, exponential, trig, etc. In your first problem, simple substitution yields 3/0 which of course is infinity. Therefore, the limit does not exist.

Your second problem is a bit more complex. When you attempt substitution and receive 0/0 as an answer (like in your second question), you can not make any judgments about the function's behavior as x approaches the limiting value. 0/0 is called an indeterminate form. There are many types of indeterminates. 0/0 INF/INF 0*INF and others. In these situations the function needs to be manipulated into a different form. There is more than one way to do this. I won't cover the more advanced techniques here. I'll only confuse you as you are just starting out. Besides, it's important to learn these basic techniques first.

Now, I used a basic technique to compute the limit for the second question. Are you familiar with rationalization? If you are studying Calculus then you should be. After all, it's an algebra concept. Rationalize the numerator by multiplying it by its conjugate (4+t)^.5 + 2 Essentially, this yields a difference of squares. Now, as not to change the fraction, you must multiply the denominator by the same entity. In the numerator, you have 4+t-4 and in the denominator, t^.5 * ((4+t)^.5 +2) In the numerator, the 4 and -4 cancel and you are left with a "t". This is what you need. All factors in the top and bottom. The t^.5 in the denominator cancels with the "t" in the numerator, and you are left with a t^.5 up top and (4+t)^.5 + 2 below. Now, re-evaluate the limit. Remember, we haven't changed a thing. We only multiplied by one, and rearranged. When you perform substitution now, you get 0/4.