# Does this limit exists ? $\lim_{x \to 0} \sec\sqrt{x}$

Does this limit exists ?

$$\lim_{x \to 0} \sec\sqrt{x}$$

Left hand limit : $\lim_{x\to 0^-} \sec\sqrt{x}$ does not exist.

Right hand limit : $\lim_{x \to 0^+} \sec \sqrt{x} = 1$.

• The LHS limit isn't well-defined, so I think it means one side limit. – John Oct 27 '14 at 15:27
• It depends how you define $\cos(x)$ when $x$ is an imaginary number. According to Wolram it is defined see here both left and right hand limits are defined both are 1 so the limit exists and is 1. – Warren Hill Oct 27 '14 at 16:18
• @warren : yes limit exists, but what is the method for that to prove.. please elaborate thanks – sultan Oct 27 '14 at 17:18

We can define $\cos(z)$ by use of the Taylor Series

When we do we find

$$\cos(z) = \sum_{k=0}^{\infty}\dfrac{(-1)^k \cdot z^{2k}}{(2k)!} \approx 1 - \dfrac{z^2}{2!}+\dfrac{z^4}{4!}-\dfrac{z^6}{6!}+\dfrac{z^8}{8!} - \dfrac{z^{10}}{10!}+\ldots$$

Now we want

$$\lim_{x \to 0}\sec\left(\sqrt{x}\right) = \lim_{x \to 0} \dfrac{1}{\cos\left(\sqrt{x}\right) }$$

Now consider both the right and left hand limits

As we reduce $x$ from some small positive value to zero it's clear we are trying to find

$$\lim_{u \to 0^+} \dfrac{1}{\cos(u)} = \dfrac{1}{1} = 1$$

Now consider small negative $x$ as we increase $x$ to zero.

Note that the square root of a negative number is imaginary so the limit we want to find is

$$\lim_{u \to 0^-} \dfrac{1}{\cos(i \ u)} = \dfrac{1}{1} = 1$$

From our definition of $\cos(z)$ using the Taylor Series above and noting $i^2 = -1$ we have

$$\lim_{u \to 0} \cos(i \ u) \approx 1 + \dfrac{u^2}{2!}-\dfrac{u^4}{4!}+\dfrac{u^6}{6!}-\dfrac{u^8}{8!} + \dfrac{u^{10}}{10!}-\ldots$$

Which we can see will be 1 when $u = 0$

We have now shown

$$lim_{x \to 0^+} \sec{\left(\sqrt{x}\right)} = 1 \text{ and } lim_{x \to 0^-} \sec{\left(\sqrt{x}\right)} = 1$$

Since both limits exist and have the same value there is no discontinuity at $x = 0$ so the limit exists and we have shown it to be 1.