Limit where ${x\to 0}$

Question

Can someone please help me with solving this limit, I do not even know how to start ... $$\lim_{x\to0}\frac{(e^{x^2}-1)\cdot\sin\left(\frac{x}{2}\right)^3}{5x^5}$$

Answer 1

Note that by the standard limits for $y\to 0$

$$\frac{\sin y}{y}\to 1 \quad \quad \frac{e^y-1}{y}\to 1$$

we obtain

$$\frac{(e^{x^2}-1)\cdot\sin({x/2})^3)}{5x^5}=\frac{(e^{x^2}-1)}{x^2}\cdot \frac{\sin({x/2})^3}{(x/2)^3}\cdot\frac{1}{40}\to1\cdot1\cdot\frac{1}{40}=\frac{1}{40}$$

Answer 2

HINT

Use the following two limits $\lim_{x \to 0} \frac{\sin x}{x} = 1$ and $\lim_{x \to 0} \frac{e^x - 1}{x} = 1$

Answer 3

$$ \left(e^{x^2}-1\right)\left(\sin\left(\frac{x}{2}\right)^3\right)=\left(x^2+o\left(x^2\right)\right)\left(\frac{x^3}{8}+o\left(x^3\right)\right) $$ Hence

$$ \frac{\left(e^{x^2}-1\right)\left(\sin\left(\displaystyle \frac{x}{2}\right)^3\right)}{5x^5} \underset{x \rightarrow 0}{\rightarrow}\frac{1}{40} $$

Answer 4

$$\lim_{x\to0}\frac{(e^{x^2}-1)\cdot\sin\left(\frac{x}{2}\right)^3}{5x^5}=\frac 1 {40}\lim_{x\to0}\frac{(e^{x^2}-1)\cdot\sin\left(\frac{x}{2}\right)^3}{x^2\frac {x3}{8}}$$ Using the fact that $\lim\limits_{x \to 0} \frac {\sin(x)} x=1$ $$\frac 1 {40}\lim_{x\to0}\frac{e^{x^2}-1}{x^2}=\frac 1 {40}\lim_{u\to0}\frac{e^{u}-1}{u}$$ Then apply l'Hospital's rule $$\frac 1 {40}\lim_{u\to0}\frac{e^{u}-1}{u}=\frac 1 {40}\lim_{u\to0}{e^{u}}=\frac 1 {40}$$

Answer 5

$$\lim_{x\to0}\frac{(e^{x^2}-1)\cdot\sin\left(\frac{x}{2}\right)^3}{5x^5}=\lim_{x\to0}\frac{(e^{x^2}-1)}{x^2} \frac {sin\left(\frac{x}{2}\right)^3}{(x/2)^3} \frac {1}{40} = (1)(1)(\frac{1}{40})=\frac {1}{40}$$