| Each post must include a specific title and description. Substitute and calculate the limit as follows.Set your post to "Resolved" when answered. Let z = 1 / x so that as x get large x approaches 0. Multiply numerator and denominator by the conjugate and simplifyįactor x out of the numerator and denominator and simplifyĪs x gets larger, the terms 1/x and 1/x 2 approach zero and the limit is ![]() Since x takes large values (infinity) then | x | = x. The second limit is easily calculated as followsįactor x 2 inside the square root and use the fact that sqrt(x 2 ) = | x |. We also use the fact that sin T / T approaches 1 when T approaches 0. We now calculate the first limit by letting T = 3t and noting that when t approaches 0 so does T. Use limit properties and theorems to rewrite the above limit as the product of two limits and a constant. Multiply numerator and denominator by 3t. The numerator approaches 5 and the denominator approaches 0 from the left hence the limit is given byįactor x 2 in the denominator and simplify.Īs x takes large values (infinity), the terms 2/x and 1/x 2 approaches 0 hence the limit isįactor x 2 in the numerator and denominator and simplify.Īs x takes large values (infinity), the terms 1/x and 1/x 2 and 3/x 2 approaches 0 hence the limit is 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. Find the limits of various functions using different methods.
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