# How do you find lim 1+1/x as x->0^+?

Jun 11, 2018

$\infty$

#### Explanation:

We may split the limit up as follows, recalling the fact that

${\lim}_{x \to a} \left[f \left(x\right) \pm g \left(x\right)\right] = {\lim}_{x \to a} f \left(x\right) \pm {\lim}_{x \to a} g \left(x\right)$

Then,

${\lim}_{x \to {0}^{+}} \left(1 + \frac{1}{x}\right) = {\lim}_{x \to {0}^{+}} 1 + {\lim}_{x \to {0}^{+}} \frac{1}{x}$

${\lim}_{x \to {0}^{+}} 1 = 1 ,$ in general, the limit to any value of a constant is simply that constant.

To determine ${\lim}_{x \to {0}^{+}} \frac{1}{x} ,$ envision dividing $\frac{1}{x}$ by smaller and smaller positive numbers, as we're approaching $0$ from the positive side:

$\frac{1}{0.1} = 10$
$\frac{1}{0.01} = 100$
$\frac{1}{0.001} = 1000$
$\frac{1}{0.0001} = 10000$

We can see as we approach $0$ from the positive side, our result gets larger and larger, that is, heads toward $\infty .$

Thus, ${\lim}_{x \to {0}^{+}} \frac{1}{x} = \infty$

Our result is

${\lim}_{x \to {0}^{+}} \left(1 + \frac{1}{x}\right) = 1 + \infty = \infty ,$ the $1$ doesn't change the fact that we're heading toward infinity, it doesn't impact the answer, $1$ is insignificant in comparison.