1. ## Limits

Given that $\displaystyle \lim_{x \to 0}~\frac{e^x-1}{x}=1$

Evaluate:

a) $\displaystyle \lim_{x \to 0}~\frac{\ln{(1+x)}}{x}$

b) $\displaystyle \lim_{x \to 0}~\frac{\ln{(1+5x)}}{x}$

I'm not sure how to do these ones?

2. Originally Posted by nzmathman
Given that $\displaystyle \lim_{x \to 0}~\frac{e^x-1}{x}=1$

Evaluate:

a) $\displaystyle \lim_{x \to 0}~\frac{\ln{(1+x)}}{x}$

b) $\displaystyle \lim_{x \to 0}~\frac{\ln{(1+5x)}}{x}$

I'm not sure how to do these ones?
Well, you can apply L'Hopital's rule. it doesn't use what we were given though...

3. I forgot to include one thing...these questions followed on from a question proving that $\displaystyle \lim_{x \to 0}~\frac{f(x)}{x}= \lim_{x \to 0}~\frac{x}{f^{-1}(x)}$ provided that f is an invertible function.

By the way we are instructed not to use that rule as it is not in the syllabus...

4. Originally Posted by nzmathman
Given that $\displaystyle \lim_{x \to 0}~\frac{e^x-1}{x}=1$

Evaluate:

a) $\displaystyle \lim_{x \to 0}~\frac{\ln{(1+x)}}{x}$

b) $\displaystyle \lim_{x \to 0}~\frac{\ln{(1+5x)}}{x}$

I'm not sure how to do these ones?
or this method

$\displaystyle \left. a \right){\text{ }}\mathop {\lim }\limits_{x \to 0} \frac{{\ln \left( {1 + x} \right)}}{x} = \mathop {\lim }\limits_{x \to 0} \ln {\left( {1 + x} \right)^{\frac{1}{x}}} = \left\{ \begin{gathered}x = \frac{1}{t} \Leftrightarrow \frac{1}{x} = t \hfill \\x \to 0,{\text{ }}t \to \infty \hfill \\ \end{gathered} \right\} =$

$\displaystyle = \mathop {\lim }\limits_{t \to \infty } \ln {\left( {1 + \frac{1}{t}} \right)^t} = \ln e = 1.$

$\displaystyle \left. b \right){\text{ }}\mathop {\lim }\limits_{x \to 0} \frac{{\ln \left( {1 + 5x} \right)}}{x} = \mathop {\lim }\limits_{x \to 0} \ln {\left( {1 + 5x} \right)^{\frac{1}{x}}} = \left\{ \begin{gathered}5x = \frac{1}{t} \Leftrightarrow \frac{1}{x} = 5t \hfill \\ x \to 0,{\text{ }}t \to \infty \hfill \\ \end{gathered} \right\} =$

$\displaystyle = \mathop {\lim }\limits_{t \to \infty } \ln {\left( {1 + \frac{1}{t}} \right)^{5t}} = 5 \cdot \mathop {\lim }\limits_{t \to \infty } \ln {\left( {1 + \frac{1}{t}} \right)^t} = 5\ln e = 5.$

5. Originally Posted by nzmathman
I forgot to include one thing...these questions followed on from a question proving that $\displaystyle \lim_{x \to 0}~\frac{f(x)}{x}= \lim_{x \to 0}~\frac{x}{f^{-1}(x)}$ provided that f is an invertible function.

By the way we are instructed not to use that rule as it is not in the syllabus...
ok, now we're getting somewhere.

note that the inverse of $\displaystyle y = \ln (1 + x)$ is $\displaystyle y = e^x - 1$

so the first limit is equal to $\displaystyle \lim_{x \to 0} \frac x{e^x - 1}$, and the rest is easy

now for the second one: what's the inverse of $\displaystyle \ln (1 + 5x)$ ?

6. I finally got it now!

I get the inverse as $\displaystyle \frac{e^x-1}{5}$ and it follows from there that the limit I need to find is equivalent to $\displaystyle 5\cdot\frac{1}{\lim_{x \to 0} \frac{e^x - 1}{x}} = 5$, and that answer is confirmed by my calculator

7. Originally Posted by nzmathman
I finally got it now!

I get the inverse as $\displaystyle \frac{e^x-1}{5}$ and it follows from there that the limit I need to find is equivalent to $\displaystyle 5\cdot\frac{1}{\lim_{x \to 0} \frac{e^x - 1}{x}} = 5$
yes

and that answer is confirmed by my calculator
and DeMath's post