Help with a limit involving exponential

Question

Find the following limit :

$$\lim_{x \to \infty}\dfrac{3^{x+1}+2^{x}}{2^{x+1}-3^x}$$

Any further links for studying this type of limits would be much appreciated

Answer 1

Just divide everything by $3^x$. You'll see the limit is the same as $$ \lim_{x \to \infty}\dfrac{3+(2/3)^{x}}{2 (2/3)^x - 1} = \frac{3+0}{0-1} = -3. $$

I'm assuming that $x \to \infty$ means $x \to +\infty$... If not, also consider $$ \lim_{x \to -\infty} \frac{2^x(3 (3/2)^x+1)}{2^x(2-(3/2)^x)}=\frac 12 $$

Answer 2

As has been already shown, for these kind of limits a good strategy is to factor out from numerator and denominator the "stronger" terms that is $3^x$ in this case

$$\lim_{x \to \infty}\dfrac{3^{x+1}+2^{x}}{2^{x+1}-3^x}=\lim_{x \to \infty} \frac{3^x}{3^x}\dfrac{3^{}+(2/3)^{x}}{2\cdot (2/3)^{x}-1}=\cdots$$

Se also the following examples:

• limit of $a_n= \frac{7^n + 6^n -n^{100}}{(7.1)^n-7^n+n^{101}}$ as n goes to infinity
• Find the limit of $\frac{3^{2n + 1} + 2^{3n + 1}}{7^{n+2} + 9^n}$ when $n$ goes to infinity
• How can i get the limit of $(10(n!)+10^n)/(n^{10}+n!)$ when $n$ goes to $\infty$?