![]() Simplify and find the limt.Īs x approaches 2 from the left then x - 2 approaches 0 from the left or x - 2 < 0. Since x approaches larger positive values (infinity) | x | = x. We first factor out 16 x 2 under the square root of the denominator and take out of the square root and rewrite the limit as Hence the l'hopital theorem is used to calculate the above limit as follows Hence by the squeezing theorem the above limit is given byĪs t approaches 0, both the numerator and denominator approach 0 and we have the 0 / 0 indeterminate form. Now as x takes larger values without bound (+infinity) both -1 / x and 1 / x approaches 0. Multiply both numerator and denominator by the conjugate of the numerator.ĭivide all terms of the above inequality by x, for x positive. Let us rewrite the limit so that it is of the indeterminate form 0/0.Īpply the l'hopital's theorem to find the limit.Īs x approaches 9, both numerator and denominator approach 0. We now use L'hopital's Rule and find the limit.Īs x gets larger x + 1 gets larger, 1/(x+1) approaches zero, e^(1/(x+1)) approaches 1 and e^(1/(x+1)) - 1 approaches 0 hence an indeterminate form: ∞ × 0 Let us rewrite the limit so that it is of the infinity/infinity indeterminate form. The limit from the right of 2 and the limit from the left of 2 are not equal therefore the given limit DOES NOT EXIST.Īs x approaches -1, cube root x + 1 approaches 0 and ln(x+1) approaches - infinity hence an indeterminate form 0 × infinity Substitute to obtain the limit from the right of 2 as follows Note that we are looking for the limit as x approaches 1 from the left ( x → 1 - means x approaches 1 by values smaller than 1). More exercises with answers are at the end of this page. Several Examples with detailed solutions are presented. On a semi-related note, just because the left-hand and right-hand limits are equal as they approach some value of x, it does not mean the function is continuous at this point.Find the limits of various functions using different methods. Otherwise, an ordinary does not exist, as seen above. Remember it is important to do this from both sides, so you must evaluate x = -0.9, x = -0.99, x = -0.999, to make sure the limit is the same as you approach x from both sides. By doing this you will quickly see that the function approaches some value, which is your limit. For example, if you want the limit as x approaches 1 but evaluating x = 1 is impossible. To fix this issue, you should sub values close to x, slowly getting closer and closer to x, and evaluate your function from both sides. Sometimes this may not be possible, as it may end up with the division of 0 for example. All you have to do is substitute the x value that you want the limit for, into your function. An easy method of finding a limit, if it exists, is the substitution method. Looking at your graph it easy to find the answer, which you have correctly said is 2.įinding a limit generally means finding what value y is for a value of x. The limit does not exist as x approaches 0.įinally, this is asking for the value of the function at x = 2. As the limits differ depending on direction, the answer should be the same as the question above. Checking your graph, we can easily see the limit as x approaches 0 from the right is -1. ![]() However, we must also check to see if the right-hand limit is the same. Using the same logic as above, we can see that the left-hand limit of the function as x approaches 0 is equal to 3. It is important to test the function from both sides of the limit. Thus, we can see that there is no limit as x approaches 2. However, as we see in the above answers, the limit as x approaches 2 is different depending on the direction. The third is asking for the limit as x approaches 2. Following the same logic but from the other direction, we again find your answer to be correct. The second asks for the right-hand limit (indicated by the plus sign) as x approaches 2. Doing this, you can clearly see you answer is correct. To find this you follow the graph of your function from the left of the curve to the right as x approaches 2. ![]() ![]() The first one is asking for the left-hand limit (indicated by the minus sign). Answering your questions from top to bottom:
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