Limit of $f(x)$ when $x$ goes to zero

Question

Let $f(x) = \frac{1 + \tan x + \sin x - \cos x}{\sin^2 x + x^3}$ . Find value of $\lim_{x \to 0} f(x)$ if it exists . I can solve it using L'Hospital's Rule and Taylor series but I'm looking for another way suing trigonometric identities .

Answer 1

\begin{align} \dfrac{1-\cos x+\sin x+\tan x}{\sin^2x+x^3} &= \dfrac{2\sin^2\frac{x}{2}+\sin x\dfrac{1+\cos x}{\cos x}}{\sin^2x+x^3} \\ &= \dfrac{2\dfrac{\sin^2\frac{x}{2}}{x^2}+\dfrac{\sin x}{x}\dfrac{1+\cos x}{x\cos x}}{\dfrac{\sin^2x}{x^2}+x} \\ &\to\dfrac{\dfrac12+1\times\infty}{1+0}=\infty \end{align} as $x\to0$.

Answer 2

$$\lim_{x\rightarrow0}\frac{1 + \tan x + \sin x - \cos x}{\sin^2 x + x^3}=\lim_{x\rightarrow0}\frac{2\sin^2\frac{x}{2} +2\sin\frac{x}{2}\cos\frac{x}{2}\left(\frac{1}{\cos{x}}+1\right) }{\sin^2 x + x^3}=$$ $$=\lim_{x\rightarrow0}\left(\frac{\sin\frac{x}{2}}{\frac{x}{2}}\cdot\frac{\sin\frac{x}{2} +\cos\frac{x}{2}\left(\frac{1}{\cos{x}}+1\right) }{\frac{\sin x}{x}\cdot\sin{x} + x^2}\right)=\infty$$ because $\frac{\sin{x}}{x}\rightarrow1$.

Answer 3

As $x\to 0$ we have

$\tan x\sim x;\;\sin x \sim x;\;1-\cos x\sim \dfrac{x^2}{2}$

So the limit becomes

$$\lim_{x\to 0} \frac{1 + \tan x + \sin x - \cos x}{\sin^2 x + x^3}=\lim_{x\to 0}\frac{x+x+\frac{x^2}{2}}{x^2+x^3}=\lim_{x\to 0}\frac{4x+x^2}{2(x^2+x^3)}=\lim_{x\to 0}\frac{x(4+x)}{2x^2(1+x)}=$$ $$=\lim_{x\to 0}\frac{4+x}{2x(1+x)}=\infty$$