# How do you evaluate this limit?

• Leo Liu
In summary: This is helpful. Thank you. I didn't expect this question could be so complex.By Taylor expansion it is\lim_{x \rightarrow +0} (\frac{x}{2}+o(x^2))^{x+o(x^3)}=\lim_{x \rightarrow +0} x^x=1-0Someone on reddit sent me his work:So the key is to use the substitution ##x=t^2## to simplify the expression and to notice ##\lim_{t\to 0} 2t/\sin t=2## before applying the product rule of limit.
Leo Liu
Homework Statement
$$\lim_{x\to 0^+}(1-\cos (\sqrt x))^{\sin(x)}$$
Relevant Equations
.
I tried taking e^ln but to no avail. Please help! Thanks.

My attempt:
$$\lim_{x\to 0^+}(1-\cos (\sqrt x))^{\sin(x)}$$
$$\lim_{x\to 0^+}e^{\ln (1-\cos\sqrt x)^{\sin x}}$$
$$\lim_{x\to 0^+}e^{{\sin x}\ln (1-\cos\sqrt x)}$$
$$\lim_{x\to 0^+}\exp(\frac{\ln (1-\cos\sqrt x)}{1/\sin x})$$
If I apply Lhospital's rule to this limit, the result will be quite complicated and will remain in intermediate form. I also reached wolfram alpha for help, yet the step by step solution is terse.

Leo Liu said:
Homework Statement:: $$\lim_{x\to 0^+}(1-\cos (\sqrt x))^{\sin(x)}$$
Relevant Equations:: .

I tried taking e^ln but to no avail. Please help! Thanks.

My attempt:
$$\lim_{x\to 0^+}(1-\cos (\sqrt x))^{\sin(x)}$$
$$\lim_{x\to 0^+}e^{\ln (1-\cos\sqrt x)^{\sin x}}$$
$$\lim_{x\to 0^+}e^{{\sin x}\ln (1-\cos\sqrt x)}$$
$$\lim_{x\to 0^+}\exp(\frac{\ln (1-\cos\sqrt x)}{1/\sin x})$$
If I apply Lhospital's rule to this limit, the result will be quite complicated and will remain in intermediate form. I also reached wolfram alpha for help, yet the step by step solution is terse.
Just keep going with l'Hopital, complicated or not.

You don't need the exponential.

Leo Liu
You could use the trig identity ##\sin^2 \frac \theta 2 = \frac{1-\cos \theta}{2}## to simplify the numerator before differentiating.

Leo Liu
PeroK said:
Just keep going with l'Hopital, complicated or not.

You don't need the exponential.
I am lazy so I let MMA do the job for me. Here is what I got after differentiating the fraction twice:

Although the graph verifies that the limit is indeed 0 as x approaches 0 from right side, it seems that Num2/Den2 is still in intermediate form. I doubt that we will obtain an answer by blindly differentiating the fraction. Thanks anyway.

Do you know what the Taylor series for sine and cosine are, or have you not reached that part in your class yet?

After the first round of differentiation, you should be able to show that you end up with
$$-\frac 14 \cdot \frac{\sin \sqrt x}{\sqrt x} \cdot \frac{\sin x}{\sin \frac{\sqrt x}2} \cdot \frac{\tan x}{\sin \frac{\sqrt x}2}.$$ Calculate the limit of each factor.

Leo Liu
vela said:
After the first round of differentiation, you should be able to show that you end up with
$$-\frac 14 \cdot \frac{\sin \sqrt x}{\sqrt x} \cdot \frac{\sin x}{\sin \frac{\sqrt x}2} \cdot \frac{\tan x}{\sin \frac{\sqrt x}2}.$$ Calculate the limit of each factor.
This is helpful. Thank you. I didn't expect this question could be so complex.

By Taylor expansion it is
$$\lim_{x \rightarrow +0} (\frac{x}{2}+o(x^2))^{x+o(x^3)}=\lim_{x \rightarrow +0} x^x=1-0$$

Last edited:
Someone on reddit sent me his work:

So the key is to use the substitution ##x=t^2## to simplify the expression and to notice ##\lim_{t\to 0} 2t/\sin t=2## before applying the product rule of limit.

## 1. What is a limit?

A limit is a mathematical concept that represents the value that a function or sequence approaches as the input or index approaches a specific value. It is used to describe the behavior of a function near a certain point or as the input approaches infinity or negative infinity.

## 2. How do you evaluate a limit algebraically?

To evaluate a limit algebraically, you can use various techniques such as direct substitution, factoring, and rationalization. You can also use limit laws and properties to simplify the expression and then substitute the value of the input to find the limit.

## 3. What is the difference between a one-sided limit and a two-sided limit?

A one-sided limit only considers the behavior of the function as the input approaches the specific value from one side (either from the left or the right). A two-sided limit, on the other hand, considers the behavior of the function as the input approaches the specific value from both sides.

## 4. When is L'Hôpital's rule used to evaluate a limit?

L'Hôpital's rule is used when evaluating limits of indeterminate forms, such as 0/0 or ∞/∞. It states that if the limit of the quotient of two functions is indeterminate, then the limit of the quotient of their derivatives will be the same.

## 5. Can a limit not exist?

Yes, a limit can fail to exist if the function has a jump or a discontinuity at the specific value of the input. It can also not exist if the function oscillates or approaches different values from the left and right sides.

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