indeterminate forms of limits

Question

1)here's a basic question if we have some situaion like $\lim_{x \to \infty} a^x$ where $a$ is $>0$ the value of this limit will eventually be infinite will this be considered to be inderminate form of a limit ? $$$$2)And why $1^\infty$ is considered a inderminant form of a limit because we already know 1 raised to any power is going to be one but still we deal with this considering it as inderminant form of a limit ? $$$$3) we have one simple problem $\lim_{x \to 0} \frac{sin\frac{1}{x}}{\frac{1}{x}}$ what will be value of limit ? or the limit does't exists

Answer 1

1) This limit is not an indeterminate form. Let $a \in ]0,+\infty[$. Since, $a^{x} = \exp(x\ln(a))$ for all $x \in \mathbb{R}$, you have :

$$ \lim \limits_{x \rightarrow +\infty} a^{x} = \begin{cases} +\infty & \text{if } a > 1 \\ 1 & \text{if } a = 1 \\ 0 & \text{if } a < 1 \\ \end{cases} $$

2) For example :

$$ \lim \limits_{x \rightarrow +\infty} \left( 1+\frac{1}{x} \right)^{x} = e $$

3) You have :

$$ \frac{\sin(\frac{1}{x})}{\frac{1}{x}} = x \sin(\frac{1}{x}) $$

and since $\sin(\frac{1}{x})$ is bounded, we have :

$$ \lim \limits_{x \rightarrow 0} \frac{\sin(\frac{1}{x})}{\frac{1}{x}} = 0 $$

Answer 2

The case of $1^\infty$ relates to limits of the form $\lim _{x\to x_0}f(x)^{g(x)}$, where $\lim_{x\to x_0} f(x) = 1$ and $\lim_{x\to x_0} g(x) = \infty$. Tese limits cannot be resolved by simply stating that they're equal to 1, since everything depends on the functions $f(x)$ and $g(x)$. For instance, it is well-known that $\lim_{x\to \infty} (1+\frac1{x})^x = e$ (finite!), however, $\lim_{x\to \infty} (1+\frac1{x})^{x^2}$ will be infinite... In your first case ( 1) ) the limit $\lim_{x\to +\infty} a^x$ exists and equals $+\infty$, and $\lim_{x\to -\infty} a^x$ also exists and equals 0 (in case $a > 1$).