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

Question

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

Answer 1

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.

Answer 2

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?

Answer 3

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

Answer 4

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

Answer 5

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 $1/5$.

Answer 6

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

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