Limits with logarithm

Question

I have the following:

$$\lim_{x\to 0} \frac{1-\frac{\sin(x)}{x}}{\log (1+2x^2)}.$$

I set $2x^2=t$ knowing that $\lim_{t\to 0} \frac{\log (1+t)}{t} =1$

but I am not sure if I am doing it right.

Answer 1

$$\displaystyle \lim_{x\rightarrow 0}\frac{x-\sin x}{x\cdot \ln(1+2x^2)} = \lim_{x\rightarrow 0}\frac{x-\sin x}{2x^3}\times \lim_{x\rightarrow 0}\frac{2x^2}{\ln(1+2x^2)} = \lim_{x\rightarrow 0}\frac{x-\sin x}{2x^3}\times 1$$

Above we have used $$\displaystyle \bullet \lim_{y\rightarrow 0}\frac{\ln(1+y)}{y} =1 $$

Now our limit convert into $$\displaystyle \lim_{x\rightarrow 0}\frac{x-\sin x}{2x^3}$$

Now Using $\bf{DL,Hopital\; Rule\; 3\; times }$

So Limit $$\displaystyle \lim_{x\rightarrow 0}\frac{1-\cos x}{6x^2}$$

So Limit $$\displaystyle \lim_{x\rightarrow 0}\frac{\sin x}{12x}$$

So Limit $$\displaystyle \lim_{x\rightarrow 0}\frac{\cos x}{12} = \frac{1}{12}$$

Answer 2

Use equivalents:

Taylor's formula at order $3$ gives: $$1-\frac{\sin x}x=1-\frac1x\Bigl(x-\frac{x^3}6+o(x^3)\Bigr)=\frac{x^2}6+o(x^2)\sim_0\frac{x^2}6.$$ Also $\ln(1+2x^2)\sim_0 2x^2$, hence $$\frac{1-\dfrac{\sin x}x}{\ln(1+2x^2)}\sim_0\frac{\dfrac{x^2}6}{2x^2}=\frac1{12}.$$