Limit of function, when $x$ approaches $a$
So w have to estimate limit of function when the x approaches a. Function is given on real numbers by following term
$f(x)= \dfrac{\ln\frac{x}{a}}{x-a}$
Without L'Hôpital's rule.
So i go with
$\lim \limits_{x \to a}\dfrac{\ln\frac{x}{a}}{x-a}=\lim \limits_{x \to a}\dfrac{\ln(x)-ln(a)}{x-a}$
I know that this limit would be just a derivative of ln(x) at point "a", but this leaves me with nothing. So now is my question... I need any clue, how to deal with this limit. I would greatly appriciate any hint.
PS. As $\ln(x)$ I understand $\log_e(x)$. Also $ln(x)$ is for me inverse function to $e^x$
$\endgroup$ 113 Answers
$\begingroup$You can write
$\lim \limits_{x \to a}\dfrac{\ln\frac{x}{a}}{x-a} = \lim \limits_{x \to a} \dfrac{\ln\frac{x}{a}}{a(\frac{x}{a} - 1)} = \frac {1} {a} \lim \limits_{x \to a} \dfrac{\ln\frac{x}{a}}{(\frac{x}{a} - 1) }$.
Now use the composition law for limits to see that
$\lim \limits_{y \to 1} \dfrac{\ln{y}} {y - 1} = 1 $ and $\frac{1} {a} \lim \limits_{y \to 1} \dfrac{\ln{y}} {y - 1} =\frac{1} {a} $
(if you're not allowed to use the last formula... prove it!)
We note that, for the logarithmic proprieties,
$\lim \limits_{y \to 1} \dfrac{\ln{y}} {y - 1} = \lim \limits_{y \to 1} \ln {y^\frac{1}{(y - 1)}}$
Now, let
$ y - 1 := \frac{1} {t} $.
It's obvious that ${ t\to \infty} $ when $ {y\to 1}$.
You're left with
$\lim \limits_{t\to \infty}\ln{(1 + \frac{1} {t})^t}$
and, using the identity
$\lim \limits_{t\to \infty} ({1 + \frac {1} {t})^t = e} $
the thesis follows from the composition law for limits.
$\endgroup$ 0 $\begingroup$If you don't want to use L'Hôpital's rule. You can think another way, try Taylor expantion for $ln(x)$ at $x=a$.
$ln(x)=ln(a)+\frac{1}{a}(x-a)+(-\frac{1}{a^2})\frac{(x-a)^2}{2}+O(x-a)$
then,
$\lim \limits_{x \to a}\dfrac{\ln\frac{x}{a}}{x-a}\\
=\lim \limits_{x \to a}\dfrac{\ln(x)-ln(a)}{x-a}\\
=\lim \limits_{x \to a}\dfrac{\frac{1}{a}(x-a)+(-\frac{1}{a^2})\frac{(x-a)^2}{2}+O(x-a)}{x-a}\\
=\frac{1}{a}+\lim \limits_{x \to a}(-\frac{1}{a}\frac{(x-a)}{2}+O(x-a))\\
=\frac{1}{a}$
Still the same answer.
$$f(x)= \dfrac{\ln\frac{x}{a}}{x-a}=\frac{1}{a}\cdot\frac{\ln\frac{x}{a}}{\frac{x}{a}-1}$$ $$\lim \limits_{t \to 1}\dfrac{\ln t}{t-1}=???$$
The tangent to $g(t)=\ln t$ at the point $t=1$ is $$\frac{y-0}{x-1}=g'(1)=\frac{1}{1}=1\iff y=x-1$$ and $g(t)$ being derivable at $t=1$ it coincides in differentials with its tangent. Hence $\lim \limits_{t \to 1}\dfrac{\ln t}{t-1}=1$. Consequently $$\lim \limits_{x \to a}\dfrac{\ln\frac{x}{a}}{x-a}=\frac{1}{a}$$
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