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

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.

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

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