Limit without l'hospital,derivatives,taylor series...just using limit properties and some basic limits

Question

As the title says i have to solve this two limits without the help of these,i only know some basic limits from 1st semester calculus:

1st limit:

$$\lim_{x\to 1}\frac{\sin{\pi x}}{\ln(2x^2-1)}$$

And the 2nd limit:

$$\lim_{x\to -2}(2\sqrt{-x-1}-1)^{1/(4^{-x}-16)}$$

Any help would be appreciated. Thank you in advance.

Edit: First one i changed a little into: $\lim_{x\to1} \frac{\sin{\pi x}(2x^2-2)x}{\ln(2x^2-1)(2x^2-2)x}$ but i encountered a problem,while i still get a 0 at the bottom and it doesn't help I know the result for first limit if it helps: $-\frac{\pi}{4}$

Answer 1

Let us consider $$A=\frac{\sin({\pi x})}{\log(2x^2-1)}$$ and make a change of veriable $x=y+1$. So, after minor rearrangements and simplifications, $$A=\frac{\sin{(\pi (y+1))}}{\log(2(y+1)^2-1)}=-\frac{\sin (\pi y)}{\log (1+2 y (y+2))}$$ Now, you know that, for small values of $z$, $\sin(z)\approx z$ and that $\log(1+z)\approx z$. So replacing each time $z$ by its appropriate value ($\pi y$ for the numerator) and $2y(y+2)$ for the denominator, we have $$A\approx -\frac{\pi y}{2y(y+2)}=-\frac{\pi}{2(y+2)} $$ and $y$ is going to $0$. So, as a final result$$\lim_{y\to 0} A=-\frac{\pi}4$$

For the second one $$B=(2\sqrt{-x-1}-1)^{\frac{1}{4^{-x}-16}}$$ make a change of veriable $x=y-2$; this makes $$B=\left(2 \sqrt{1-y}-1\right)^{\frac{1}{4^{2-y}-16}}$$ So, $$\log(B)={\frac{1}{4^{2-y}-16}}\log\left(2 \sqrt{1-y}-1\right)$$ Using the same trick as before $$\log\left(2 \sqrt{1-y}-1\right)\approx \log(1-y)\approx -y$$ since $\sqrt{1+z}\approx \frac z 2$. Now consider $${4^{2-y}-16}=16(4^{-y}-1)=16(e^{-y \log(4)}-1)$$ and you know thet close to $0$, $e^{az}\approx 1+az$ which makes $$16(e^{-y \log(4)}-1)\approx -16 y\log(4)$$ All of that makes $$\log(B)\approx \frac y {16 y\log(4)}=\frac 1 {16 \log(4)}$$ and then $$B\approx e^{\frac{1}{16 \log (4)}}$$

Answer 2

Let's assume you know that

$$\lim_{u\to0}{\sin u\over u}=1\quad\text{and}\quad \lim_{u\to0}{\ln u\over u-1}=1$$

Then the key to the first limit is to use the general theorem

$$\lim_{x\to c}{f(x)\over g(x)}={\lim_{x\to c}f(x)\over\lim_{x\to c}g(x)}$$

(provided $\lim_{x\to c}g(x)\not=0$) applied to

$$f(x)={\sin\pi x\over x-1}\quad\text{and}\quad g(x)={\ln(2x^2-1)\over x-1}$$

Using the trig identity $\sin(\pi x-\pi)=-\sin\pi x$, we see that

$$\lim_{x\to1}{\sin\pi x\over x-1}=\lim_{x\to1}{-\pi\sin\pi(x-1)\over\pi(x-1)}=-\pi\lim_{u\to0}{\sin u\over u}=-\pi$$

Using the related general theorem $\lim_{x\to c}(f(x)g(x))=\lim_{x\to c}f(x)\cdot\lim_{x\to c}g(x)$, we see that

$$\lim_{x\to1}{\ln(2x^2-1)\over x-1}=\lim_{x\to1}\left({\ln(2x^2-1)\over(2x^2-1)-1}\cdot{(2x^2-1)-1\over x-1}\right)=\lim_{u\to0}{\ln u\over u-1}\cdot\lim_{x\to1}2(x+1)=4$$

Putting these together gives $-\pi/4$ for the first limit.

Something similar should work for the second limit (after first taking the logarithm). The trick is to identify the basic limits you're allowed to assume, and then tease things apart algebraically, using general theorems about limits.