How to Evaluate a Particular Right Hand Limit?

Question

How do we rigorously evaluate the limit $$ \lim_{x \to 0^{+}} \frac{0.77-0.3x-0.2^x}{0.7x}? $$ It is not hard to see it is going to be $-\infty$ by numerical calculation but how do we fix with logical arguments?

Thank you!

Answer 1

Since $\lim_{x\to0^+}0.77-0.3x-0.2^x=0.77-1=-0.23<0$, since $\lim_{x\to0^+}0.7x=0$ and since $0.7x>0$ when $x>0$,$$\lim_{x\to0^+}\frac{0.77-0.3x-0.2^x}{0.7x}=-\infty.$$

Answer 2

$$\begin{eqnarray*} L &=& \lim_{x \to 0^{+}} \frac{0.77-0.3x-0.2^x}{0.7x} &=& \lim_{x \to 0^{+}} \frac{0.77 - 0.2^x}{0.7x} - \lim_{x \to 0^{+}} \frac{0.3x}{0.7x}\\ &=& \lim_{x \to 0^{+}} \frac{0.77 - 0.2^x}{0.7x} - \frac{3}{7} \end{eqnarray*} $$

For $x$ small enough, $0.77-0.2^x$ shall be negative, for $0.2^x$ will be close to $1$. However, the denominator $0.7x$ will be positive. Indeed,

$$\frac{0.77 - 0.2^x}{0.7x} \to_{x\to 0+} \frac{-0.23^+}{0^+}, $$

which converges to $-\infty$. So $L$ = $-\infty - 3/7 = -\infty$.

Answer 3

Rewrite the limit and replace $\frac{x}{x}$ by 1 (as $x$ never gets to $0$) as follows:

$$ \lim_{x \to 0^{+}} \left[ - \frac{3}{7} + \left(\frac{.77 - .2^{x}}{.7x}\right) \right] = \lim_{x \to 0^{+}} \left(-\frac{3}{7} + \frac{-.23}{.7x}\right) = - \infty$$