# Thread: limits using L'Hopital's rule or otherwise and using leading behaviours to sketch

1. ## limits using L'Hopital's rule or otherwise and using leading behaviours to sketch

Hi, I tried to do the following problems but I didn't get very far.

Find each limit using L'Hopital's rule (if possible; check assumptions before you try to use it!) or otherwise.

lim (tanx/x) as x approaches 0
-->since it's indeterminate form : 0/0, i used the rule and got 0 but the answer is 1.

lim (x^(4x)) as x approaches 0 from the right side
-->i tried to use the rule for it by changing it into a fraction: 1/ x^(-4x) but then when i differentiate it's almost the same thing

a(c) = 5c / e^(2c)
-->how do i find the leading behaviours if there is only one term in the numerator and in the denominator?

thanks!

2. ## Re: limits using L'Hopital's rule or otherwise and using leading behaviours to sketc

$\displaystyle \lim_{x \to 0}\frac{\tan x}{x}=\lim_{x \to 0}\frac{\sec ^{2}x}{1}=1$

3. ## Re: limits using L'Hopital's rule or otherwise and using leading behaviours to sketc

$\displaystyle \lim_{x \to 0+}x^{4x}=\lim_{x \to 0+}e^{ln(x)4x}$
$\displaystyle =e^{\lim_{x \to 0+}ln(x)4x}$
$\displaystyle =e^{4\lim_{x \to 0+}\frac{ln(x)}{\frac{1}{x}}}$
$\displaystyle =1$

4. ## Re: limits using L'Hopital's rule or otherwise and using leading behaviours to sketc

An other way to solve the first limit:
$\displaystyle \lim_{x \to 0} \frac{\tan(x)}{x}=\lim_{x \to 0} \frac{\frac{\sin(x)}{\cos(x)}}{x}=\lim_{x\to 0} \left(\frac{\sin(x)}{x}\cdot \frac{1}{\cos(x)}\right)=\lim_{x\to 0} \frac{\sin(x)}{x} \cdot \lim_{x \to 0}\frac{1}{\cos(x)}=1.1=1$