Need help solving these limits

Question

$$1. \lim_{n \to \infty} \left(\frac{2n^2 + 5}{1 + 5 + ... + (4n - 3)} + \frac{5}{n}\right)$$ $$2. \lim_{x \to -1} \left(\frac{x^4 + 2x^3 + 4x^2 + 6x + 3}{x^3 - 3x^2 - 9x - 5}\right)$$ $$3. \lim_{x \to \infty} \left(\frac{x-1}{14x+5}\right)^{1-3x}$$ $$4. \lim_{x \to 0} (1+\sin{(7x)})^{(\sqrt[\leftroot{-2}\uproot{2}5]{1-9x+x^2}-1)^{-2}}$$ $$5. \lim_{x \to 0} \frac{x \cdot \ln(1+4x^2)}{(1-2x^3)^8-1}$$

Answer 1

HINTS:

• 1)

$$\lim_{n\to\infty}\left(\frac{2n^2+5}{1+5+\dots+(4n-3)}+\frac{5}{n}\right)=$$ $$\lim_{n\to\infty}\left(\frac{2n^2+5}{\sum_{k=1}^{n}(4k-3)}+\frac{5}{n}\right)=$$ $$\lim_{n\to\infty}\left(\frac{2n^2+5}{2n^2-n}+\frac{5}{n}\right)=$$ $$\lim_{n\to\infty}\frac{2n^2+5}{2n^2-n}+\lim_{n\to\infty}\frac{5}{n}=$$ $$\lim_{n\to\infty}\frac{2n^2+5}{2n^2-n}+0=$$ $$\lim_{n\to\infty}\frac{2n^2+5}{2n^2-n}=$$ $$\lim_{n\to\infty}\frac{2+\frac{5}{n^2}}{2-\frac{1}{n}}=$$ $$\frac{2+0}{2-0}=\frac{2}{2}=1$$

• 2) $$\lim_{x\to-1}\frac{x^4+2x^3+4x^2+6x+3}{x^3-3x^2-9x-5}=\lim_{x\to-1}\frac{x^2+3}{x-5}=\frac{(-1)^2+3}{(-1)-5}=-\frac{2}{3}$$

• 3) $$\text{L}=\lim_{x\to\infty}\left(\frac{x-1}{14x+5}\right)^{1-3x}=\exp\left(\lim_{x\to\infty}(1-3x)\lim_{x\to\infty}\ln\left(\frac{x-1}{14x+5}\right)\right)$$

$$\text{so when}\space\space x\to\infty\space\space\text{then}\space\space\text{L}\to\infty$$

• 4) Notice: a two-sided limit does not exist!

$$\lim_{x\to0^-}\left(1+\sin(7x)\right)^{\frac{1}{\left(\sqrt[5]{x^2-9x+1}-1\right)^2}}=\exp\left(\lim_{x\to0^-}\frac{\ln(1+\sin(7x))}{\left(\sqrt[5]{x^2-9x+1}-1\right)^2}\right)=0$$

Answer 2

HINT: for 1) prove by induction that $$\sum_{i=1}^n 4i-3=2n(n+3)$$ for 2) show that $$\frac{x^4+2x^3+4x^2+6x+3}{x^3-3x^2-9x-5}={\frac {{x}^{2}+3}{-5+x}}$$ for 3) the limit doesn't exist
for 4) the limit doesn't exist
for 5) prove by L'Hospital that the limit is given by $$-\frac{1}{4}$$