Limits , indeterminate form

Question

$$(\lim_{x\to 0} (\cos2x)^{\frac 2{x^2}} = e^{-6})$$ My doubt is that why it's indeterminate? We can see $x \neq 0$, i.e., it is tending to $0$. So $\cos2x$ cannot give a value equal to $1$ or greater than $1$ but the answer is $e^{-6}$ and I even checked on desmos it's correct. But according to me, it should be zero. Where am I wrong? Please help me out. I am a high school student so please answer accordingly. Thank you

Answer 1

You have this: $$\lim_{x\to 0} (\cos(2x))^{2/x^2}$$

Note that $\lim_{x\to 0} \cos(2x) = \cos(0) = 1.$ On the other hand, $\lim_{x\to 0}\frac{2}{x^2} = \infty.$ So we have something that is close to $1$ (but less than $1$) being raised to a large power, which gives us a result that is not so close to $1.$ For example, if we try $x = \frac{1}{100},$ we have $\cos(2x) = \cos\left(\frac{2}{100}\right)\approx 0.9998,$ which is close to $1,$ but then we have to raise this to the power $\frac{2}{(1/100)^2} = 20000,$ and the result is approximately $$ 0.9998^{20000} \approx 0.01831, $$ which is not nearly as close to $1$ as $0.9998$ is.

If the exponent were increasing to $\infty$ but not as quickly as this, the limit would be closer to $1$ (or maybe even equal to $1$). If the exponent were increasing a lot faster than this, the limit would be even smaller, perhaps even equal to zero. That's what makes this an indeterminate form; we don't know just by looking at the limits of the two pieces (the base and exponent) what the result will be.

The exact place where this limit happened to end up was determined by how fast the exponent grew vs. how fast $\cos(2x)$ approached $1.$ The details are given in other answers.

Answer 2

Hint: $$(\cos2x)^{1/2x^2}=\left((1-2\sin^2x)^{1/(-2\sin^2x)}\right)^{(-\sin^2x/x^2)}$$

Answer 3

Hint: $\cos(2x)^{\dfrac{2}{x^2}} = \exp(\dfrac{2}{x^2}\log(\cos2x))$

P.s. Something is wrong with your question and your answer, check it, please.

Answer 4

Note that

• $\cos 2x\to 1$

• $\frac 2 {x^2}\to \infty$

thus it is an indeterminate form $1^{\infty}$.

Then recall that

• $\cos 2x=1-2\sin^2 x$

and then using $A^B=e^{B\log A}$

$$(\cos2x)^{1/2x^2}=e^{\frac2{x^2}\log(1-2\sin^2x)}=e^{\frac{\sin^2x}{x^2}\log\left[(1-2\sin^2x)^{\frac 1{2\sin^2x}}\right]}\to \frac 1 e$$

indeed by standard limits

• $\log\left[(1-2\sin^2x)^{\frac 1{2\sin^2x}}\right]\to \log (e^{-1})=-1$
• $\frac{\sin^2x}{x^2}\to 1$