Can we use Binomial Approximations when evaluating limits?

Question

I came across this question,

Compute $$L=\lim\limits_{x\to 0}{\frac{\sqrt[3]{1+\sin^2 x} \hspace{2mm}-\sqrt[4]{1-2 \tan x}}{\sin x + \tan^2 x}}$$

Can I use Binomial Approximations here?

As $x\to 0 \implies \sin^2 x\to 0$

$\therefore \left(1+\sin^2 x \right)^\tfrac{1}{3} \approx 1+\frac{sin^2 x}{3}$

Similarly, $\left (1-2 \tan x \right)^\tfrac{1}{4} \approx 1-\frac{\tan x}{2}$

Now, computing limit $L$ becomes very easy and after computation.

$$\boxed{L=\frac{1}{2}}$$

And I cross checked it using a graphing calculator and the above $L$ is correct.

So, is it okay to compute limits this waya if I have no options left (excluding L'Hôpital's rule)?

Answer 1

By Taylor's theorem, $$(1+u)^{1/k} = 1 + \frac{1}{k}u + h(u) u,\qquad h(u) \to 0 \text{ as } u \to 0.$$

If the $h(u) u$ term is small enough to not matter in the rest of the calculation, then it is ok to use $1+\frac{1}{k} u$. Otherwise you may need to take a higher-order Taylor polynomial (i.e. more terms of the "binomial expansion").

In this case, keeping track of the remainder terms gives $$\frac{\frac{1}{3} \sin^2 x - \frac{1}{2} \tan x}{\sin x + \tan^2 x} + \frac{h(\sin^2 x) \sin^2 x + \tilde{h}(\tan x) \tan x}{\sin x + \tan^2 x}.$$ You showed the first term tends to $1/2$. For the second term, note that similarly $$\lim_{x\to 0} \frac{h(\sin^2 x) \sin^2 x + \tilde{h}(\tan x) \tan x}{\sin x + \tan^2 x} = \lim_{x \to 0} h(\sin^2 x) \lim_{x \to 0} \frac{\sin^2 x}{\sin x + \tan^2 x} + \lim_{x \to 0} \tilde{h}(\tan x) \lim_{x \to 0} \frac{\tan x}{\sin x + \tan^2 x} = 0 \cdot 0 + 0 \cdot 1 = 0.$$

Answer 2

$$\lim_{\lim_{x\to0}}\dfrac{(1+\sin^2x)^{1/3}-(1-2\tan x)^{1/4}}{\sin x+\tan^2x}=\lim_{x\to0}\dfrac{(1+\sin^2x)^{1/3}-1}{\sin x+\tan^2x}-\dfrac{(1-2\tan x)^{1/4}-1}{\sin x+\tan^2x}$$

Now rationalize the numerator:

$$\lim_{x\to0}\dfrac{(1+\sin^2x)^{1/3}-1}{\sin x+\tan^2x}=\lim_{x\to0}\dfrac{(1+\sin^2x)-1}{(\sin x+\tan^2x)}\cdot\lim_{x\to0}\dfrac1{(1+\sin^2x)^{2/3}+(1+\sin^2x)^{1/3}+1}$$

Finally as $x\to0\implies\sin x\to0\implies\sin x\ne0$ divide numerator & denominator by $\sin x$ $$\lim_{x\to0}\dfrac{(1+\sin^2x)-1}{(\sin x+\tan^2x)}=\lim_{x\to0}\dfrac{\sin x}{1+\tan x\sec x}=?$$

Try rationalizing the numerator: $$\lim_{x\to0}\dfrac{(1-2\tan x)^{1/4}-1}{\sin x+\tan^2x}$$