# Calculate the limit without using L'Hopital's Rule: $\lim\limits_{x \to 0} \frac{5x - e^{2x}+1}{3x +3e^{4x}-3}$

Need to calculate the following limit without using L'Hopital's Rule:

$$\lim_{x \to 0} \frac{5x - e^{2x}+1}{3x +3e^{4x}-3}$$

The problem I'm facing is that no matter what I do I still get expression of the form $$\frac{0}{0}$$ I thought maybe to use $$t = e^{2x}$$ But I still can't simplify it enough..

Thank you

• Is this a homework question? – samis Jan 6 '17 at 17:57

HINT:

As $x\to0,x\ne0$ so safely divide numerator & denominator by $x$

and use

$$\lim_{h\to0}\dfrac{e^h-1}h=1$$

Observe that the exponent of $e,$ the limit variable and the denominator are same.

• Thank you. Solved it after dividing by x. – Noam Jan 6 '17 at 10:19

$$\lim_{x \to 0} \frac{5x - e^{2x}+1}{3x +3e^{4x}-3}=\\ \lim_{x \to 0} \frac{5x - (1+2x+\frac{(2x)^2}{2!}+o(x^3))+1}{3x +3(1+4x+\frac{(4x)^2}{2!}+o(x^3))-3} =\\ \lim_{x \to 0} \frac{3x-\frac{(2x)^2}{2!}-o(x^3)}{15x +3\frac{(4x)^2}{2!}+o(x^3)} = \lim_{x \to 0} \frac{3x}{15x } =\frac{3}{15}$$

• $o(x^3)$ is wrong, this is $o(x^2)$ ! Plus no need to go to second order. why complicate things. – zwim Jan 6 '17 at 10:10

Using Taylor's series:

$$\frac{5x - e^{2x}+1}{3x +3e^{4x}-3} = \frac{5x - (1+2x + O(x^2))+1}{3x +3(1+4x+O(x^2))-3} = \frac{3x - O(x^2)}{15x+ O(x^2)} .$$

Thus, $$\lim_{x \rightarrow 0} \frac{3x - O(x^2)}{15x+ O(x^2)} = \frac{1}{5}.$$

• same remark than for Koshrotash, $o(x^2)$ is wrong. – zwim Jan 6 '17 at 10:14
• Why is it wrong? – Alex Silva Jan 6 '17 at 10:14
• because $e^x=1+x+x^2/2+o(x^2)$. – zwim Jan 6 '17 at 10:15
• @zwim Do you know the difference between $o(x)$ and $O(x)$? Here, $O(x^2)$ is correct. – xen Jan 6 '17 at 10:22
• I know they are different techniques; I was saying that your approach "starts to look like" the derivation of L'Hopital. And I am aware that all the answers are (implicitly or not) using the derivative concept. But rather than putting the same comment under every answer, I thought I would vent just once. Don't take it personally - I think it's a problem with the question, not your answer in particular. – Floris Jan 6 '17 at 20:16

Do you know the Taylor formula? If so, then $$e^x = 1 + x + o(x)$$ as $x \to 0$. Here, $o(x)$ is a function such that $o(x)/x \to 0$ as $x \to 0$. Hence $$\frac{5x - e^{2x}+1}{3x +3e^{4x}-3} = \frac{5x - (1 + 2x + o(x)) + 1}{3x + 3(1 + 4x + o(x))-3} = \frac{3x+ o(x)}{15x+o(x)} = \frac{3+o(x)/x}{15+o(x)/x}$$ which shows that your limit is $1/5$.

• $o(x)=xo(1)$, what is this ugly $o(x)/x$. – zwim Jan 6 '17 at 10:12
• @zwim This is quite standard notation. I do not see any "ugliness" of $\frac{o(x)}{x}$. – xen Jan 6 '17 at 10:24
• I prefer $o(1)$. – zwim Jan 6 '17 at 10:25

Hint: We can divide by $x$ to get $$\lim_{x \to 0} \frac {5-(\frac {e^{2x}-1}{2x}(2))}{3+3 (\frac {e^{4x}-1}{4x}(4))}$$ Can you take it from here?

• Look at lab bhattacharjee's answer. It is exactly what he said. There is a typo in your answer. – Behrouz Maleki Jan 6 '17 at 10:09

$$\frac{5x - e^{2x}+1}{3x +3e^{4x}-3}=\frac12\frac{\dfrac52-\dfrac{e^{2x}-1}{2x}}{\dfrac34+3\dfrac{e^{4x}-1}{4x}}\to\frac12\frac{\dfrac52-L}{\dfrac34+3L}.$$ (by a scaling of $x$, the two ratios tend to $L$).

Then you can take for granted that $L=1$, giving the answer $\dfrac15$.