Trig limit $\lim_{x \to 0} \frac{\tan(3x^2)+\sin^2(5x)}{x^2}$ without l’hopital rule

Question

$$\lim_{x \to 0} \frac{\tan(3x^2)+\sin^2(5x)}{x^2}$$

How can I solve this without l’hopitals rule? I can’t figure out how to simplify it because of the differing constants inside the trig functions.

Answer 1

I thought it might be instructive to present an approach the goes back to basics and relies on only elementary inequalities from geometry and the squeeze theorem. To that end, we proceed.

In THIS ANSWER, I appealed to elementary geometry and straightforward analysis to establish the inequalities

$$|x\cos(x)|\le |\sin(x)|\le |x| \tag 1$$

and

$$|x|\le |\tan(x)|\le \left|\frac{x}{\cos(x)}\right| \tag 2$$

From $(1)$, we can write

$$\left|25x^2\cos^2(5x)\right|\le \left|\frac{\sin^2(5x)}{x^2}\right|\le \left|\frac{25x^2}{x^2}\right| \tag 3$$

and from $(2)$, we can write

$$\left|\frac{3x^2}{x^2}\right|\le \left|\frac{\tan(3x^2)}{x^2}\right|\le \left|\frac{\frac{3x^2}{\cos(3x^2)}}{x^2}\right| \tag 4$$

Applying the squeeze theorem to $(3)$ and $(4)$, we find that

$$\lim_{x\to 0 }\frac{\sin^2(5x)}{x^2}=25 \tag 5$$

$$\lim_{x\to 0}\frac{\tan(3x^2)}{x^2}=3 \tag 6$$

Putting $(5)$ and $(6)$ together yields the coveted limit

$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 0}\left( \frac{\tan(3x^2)+\sin^2(5x)}{x^2} \right)=28}$$

Answer 2

We have after breaking, $\lim_{x \to 0}3\frac{\tan 3x^2}{3x^2}+\{\lim_{x \to 0}5\frac{\sin 5x}{5x}\}^2=3+5^2=28$.

What I used is a standard result $\lim_{x \to 0}\frac{\sin\theta}{\theta}=1$, this directly implies $\lim_{x \to 0}\frac{\tan\theta}{\theta}=1$.

Answer 3

Both tan and sin satisfy $f(x) = x + O(x^3)$ as $x \to 0$.

Therefore, as $x \to 0$,

$\begin{array}\\ \frac{\tan(3x^2)+\sin^2(5x)}{x^2} &=\frac{3x^2+O(x^6)+(5x+O(x^3))^3}{x^2}\\ &=\frac{3x^2+O(x^6)+(5x)^2(1+O(x^2))^2}{x^2}\\ &=3+O(x^4)+25(1+O(x^2))\\ &=28\\ \end{array} $

Answer 4

You could go one step further writing $$A=\frac{\tan(3x^2)+\sin^2(5x)}{x^2}=\frac{2\tan(3x^2)-\cos(10x)+1}{2x^2}$$ Now, use the following Taylor expansions around $y=0$ $$\tan(y)=y+\frac{y^3}{3}+\frac{2 y^5}{15}+O\left(y^6\right)$$ $$\cos(y)=1-\frac{y^2}{2}+\frac{y^4}{24}+O\left(y^6\right)$$ Replacing $y$ by $3x^2$ for the tangent and $y$ by $10x$ for the cosine leads to $$A=\frac{2\left(3 x^2+9 x^6+\frac{162 x^{10}}{5}+O\left(x^{11}\right)\right)-\left(1-50 x^2+\frac{1250 x^4}{3}+O\left(x^5\right)\right)+1} {2x^2}$$ $$A=28-\frac{625 x^2}{3}+O\left(x^4\right)$$ which shows the limit and how it is approached.

Answer 5

Shorter with equivalents:

• $\tan u\sim_0u$, hence $\;\dfrac{\tan(3x^2)}{x^2}\sim_0\dfrac{3x^2}{x^2}=3$.

• $\sin u\sim_0 u$, hence $\;\dfrac{\sin^2(5x)}{x^2}\sim_0\dfrac{(5x)^2}{x^2}=25$.

  $$\llap{\text{Thus}}\hspace4em\dfrac{\tan(3x^2)+\sin^2(5x)}{x^2}=\dfrac{\tan(3x^2)}{x^2}+\dfrac{\sin^2(5x)}{x^2}\to 3+25=28.\hspace{4em}$$