# Thread: help with two limits

1. ## help with two limits

Hi

I have two limits i would like some help with

first one:

$\lim \limits_{x \to 0}\frac{10^x-e^x}{x}$

and second one

$\lim \limits_{x \to \infty} x(2tan^{-1}x-\pi)$

The first limit i thought could be rewritten like this $$\frac{e^{x*ln10}-e^x}{x}$$
which would then be simplified down to:

$\frac{10e^x-e^x}{x}$

and then using L'Hopital rule to aquire 9

The last one i don't know how to start

2. 1) $\lim_{x\to 0}\frac{e^x\left(\left(\frac{10}{e}\right)^x-1\right)}{x}=$

$=\lim_{x\to 0}e^x\cdot\lim_{x\to 0}\frac{\left(\frac{10}{e}\right)^x-1}{x}=1\cdot ln\frac{10}{e}$

because $\lim_{x\to 0}\frac{a^x-1}{x}=\ln a$

2) $\lim_{x\to\infty}\frac{2\arctan x-\pi}{\frac{1}{x}}$

and use l'Hospital.

3. How you prove that this term exist ?

$\lim_{x\to 0}\frac{a^x-1}{x}=\ln a$

4. Originally Posted by gilyos
How you prove that this term exist ?

$\lim_{x\to 0}\frac{a^x-1}{x}=\ln a$
It is well known that $\lim_{x\to 0}(1+x)^{\frac{1}{x}}=e$

Then $\lim_{x\to 0}\ln(1+x)^{\frac{1}{x}}=\lim_{x\to 0}\frac{\ln(1+x)}{x}=1$.

Now let $a^x-1=y\Rightarrow x=\log_a(1+y)$ and $y\to 0$

$\lim_{x\to 0}\frac{a^x-1}{x}=\lim_{y\to 0}\frac{y}{\log_a(1+y)}=\lim_{y\to 0}\frac{y}{\ln(1+y)}\cdot\ln a=$

$=\lim_{y\to 0}\frac{1}{\ln(1+y)^{\frac{1}{y}}}\cdot\ln a=\ln a$

5. Going off of red dog's initial response we see that $\lim_{x\to\infty}\frac{2\arctan(x)-\pi}{\frac{1}{x}}$ may be transformed by letting $z=\frac{1}{x}$ into $\lim_{z\to0}\frac{2\arctan\left(\tfrac{1}{z}\right )-\pi}{z}$ but $\arctan\left(\tfrac{1}{z}\right)=\frac{\pi}{2}-\arctan(z)$ were it readily follows that our limit is equal to $-2\lim_{z\to0}\frac{\arctan(z)}{z}$ where it follows from the definition of the derivative that our answer is $-2\cdot\frac{1}{1+0^2}=-2$.

6. Hm, sorry i didn't follow everything.

How did you know to divide $(2~arctan ~x -\pi)$ by $1/x$

7. Originally Posted by Jones
Hm, sorry i didn't follow everything.

How did you know to divide $(2~arctan ~x -\pi)$ by $1/x$
I followed what red dog suggested