Find the limits in Exercises 1–6.
3. lim(x→0⁺) (cox(√x))...

Question

Find the limits in Exercises 1–6.

3. lim(x→0⁺) (cox(√x))^(1/x)

Accepted Answer

Hello. In this video we are going to be evaluating the following limit. The limit given to us is the limit, as x approaches 0 from the right, of the function cosine of 3 square root of x, all reach the power of 1 divided by x. Now, here, we see that we have an exponential value that involves a variable of X, and what we want to do is you want to try to remove this variable of X out of the exponential exponential value. So what we can go ahead and do is we can use our natural logarithm properties, but in order to do this, we first want to set this limit equal to the generic limit L. Then what we can go ahead and do is take the natural logarithm of both sides of this equation. This will allow us to rewrite the limit as the limit as x approaches zero from the right of the natural logarithm of cosine 3 square root of x, this entire thing raised to the power of 1 divided by x, and this will equal to the natural logarithm of our generic limit l. Now by using the power rule for the natural logarithm, we can move the one divided by X to the front of the natural logarithm as a coefficient. So that will leave us with the limit, as X approaches 0 from the right, of 1 divided by X multiplied by the natural logarithm of cosine, 3 square root of X, and this will equal to our generic limit, the natural logarithm of L. Now, if we were to plug in the limit directly into the left hand side of the equation, we will end up with a quantity of 0 divided by 0. This is called an indeterminate form, and whenever we have an indeterminate form, what we want to do is we want to simplify by using Lahopito's law. So the first thing we're going to do is we're going to focus on the left hand side of this equation, and we are going to combine the terms on the left hand side of the equation together. So that is going to give us the limit, as X approaches 0 from the right, of the natural logarithm, cosine of 32 root of X, and this will all be divided by X. Now Lahopito's law tells us to take the derivatives of the numerator and denominator independently. Now, if we take the derivative of the denominator, we will just get one, so we will no longer have a denominator by taking its derivative, but for the numerator, we have to use the chain rule twice. So we will take the derivative of the natural logarithm first. The derivative of the natural logarithm is going to be 1 divided by cosine of 3 square root of x. Then we want to multiply this by the second inner function, which is going to be the derivative of cosine. That is going to give us negative sine of 3 square root of x. And then we have to apply the chamber one more time by multiplying this by the derivative of 32 root of X. That is going to give us 3 divided by 2 square root of x. Now, if we were to combine like terms and simplify, we can further simplify this limit to be the limit as X approaches zero from the right of -3 divided by 2 square root of X. Multiplied by tangent of 3 square root of x. Now, if we were to apply the limit one more time, we would end up with another 0 divided by 0 quantity, which is another indeterminate form. So again, we want to go ahead and apply Lahopito's law. Now we can further simplify the limit by applying a substitution. If we allow y to equal to the square root of x, then y has to approach 0 from the right. So by applying the substitution, we can rewrite the limit as the limit, as y approaches 0 from the right of 3. -3. Tangent of 3 y. Divided by 2 square root, or sorry, not square root anymore, just 2 Y. And if we were to move the coefficient of -3 halves to the front of the limit, we have 3 halves multiplied by the limit as y approaches 0 from the right of tangent of 3 y divided by y. Now if we were to take the derivative of the numerator and denominator independently, this will leave us with -3 halves. Multiplied by the limit as y approaches zero from the right of the derivative of tangent of 3 y, which is 3 sequin squared of 3 y divided by the derivative of y, which is 1. Now if we were to apply the limit here, this will simplify to be -3 halves multiplied by 3. And that is going to leave us with -9 halves. So this is going to be the simplified form of the left-hand side of the equation. So what we are left with is our generic limit from earlier, the natural logarithm of Y, or sorry, the natural logarithm of L. Equal to -9 halves. All we have to do now is just solve for our generic limit L, but we have to remove it from the natural logarithm. So what we will do is we will exponentiate both sides of this equation, and that will leave us with L is equal toe raised the power of -9 halves. This is going to be the value of the given limit, and the overall solution to this problem is going to be option B. So I hope this video helps you in understanding on how to approach this problem, and we will go ahead and see you all in the next video.

Find the limits in Exercises 1–6.3. lim(x→0⁺) (cox(√x))^(1/x)

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Find the limits in Exercises 1–6.
3. lim(x→0⁺) (cox(√x))^(1/x)