calculate the limit of the following function.

Question

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

For some reason I can't find the trick to solve this. Tried a lot but it always come to a place where its $\frac{0}{0}$ or $\frac{\infty}{\infty}$.

Answer 1

L'hopitals Theorem:

for $ \lim_{x\to a} $ $ \frac {f(x)}{g(x)} $ , if $ \lim_{x\to a} $ $ \frac{f(x)}{g(x)} $ =$ \frac 00 $ or $ \lim_{x\to a} $ $ \frac{f(x)}{g(x)} $ = $ \frac{-\infty}{-\infty} $ or $ \lim_{x\to a} $ $ \frac{f(x)}{g(x)} $ = $ \frac{\infty}{\infty} $

then $ \lim_{x\to a} $ $ \frac {f(x)}{g(x)} $ = $ \lim_{x\to a} $ $ \frac {f'(x)}{g'(x)} $

in your case you have a=0.

$ \lim_{x\to 0} $ = $ \frac{5*0 -e^{2*0} +1}{3*0+3e^{4*0} -3} $ = $ \frac 00 $

so as you can see you can use L'hopitals Theorem here, what it basically says is that if you get $ \frac 00 $ or $ \frac{ +-\infty}{+-\infty} $ when plugging the limit into the function, you can then differentiate the numerator and denominator separativly and get a new function

$ f(x)= 5x - e^{2x} +1 $ numerator of original function

$\frac{d}{dx} $f(x) = $5 -2e^{2x} = f'(x)$ new numerator

$ g(x)= 3x +3e^{4x} -3 $ denominator of original function

$\frac{d}{dx} $g(x) = $ 3+ 12e^{4x} =g'(x)$ new denominator

this gives you: $ \lim_{x\to 0} $ $\frac{f'(x)}{g'(x)} $ = $ \lim_{x\to 0} $ $ \frac{5-2e^{2x}}{3+12e^{4x}} $ = $ \frac{5-2e^{2*0}}{3+12e^{4*0}} = \frac {3}{15}= \frac {1}{5} $ and this is the answer

Answer 2

Hint: Multiply and divide by $x$; can you think of a good way to interpret the different parts of:

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

Also, a note on terminology: This is not a rational function, since neither the numerator nor denominator are polynomials.

Answer 3

$$\lim_{x\to 0}\frac{5x-e^{2x}+1}{3x+3e^{4x}-3}=\lim_{x\to 0}\left(\frac{1-e^{2x}}{-3+3e^{4x}+3x}+\frac{5x}{-3+3e^{4x}+3x}\right)=$$ $$\lim_{x\to 0}\frac{1-e^{2x}}{-3+3e^{4x}+3x}+5\left(\lim_{x\to 0}\frac{x}{-3+3e^{4x}+3x}\right)=$$ $$\lim_{x\to 0}\frac{1-e^{2x}}{-3+3e^{4x}+3x}+5\left(\lim_{x\to 0}\frac{\frac{\text{d}}{\text{d}x}\left(x\right)}{\frac{\text{d}}{\text{d}x}\left(-3+3e^{4x}+3x\right)}\right)=$$ $$\lim_{x\to 0}\frac{1-e^{2x}}{-3+3e^{4x}+3x}+5\left(\lim_{x\to 0}\frac{1}{3+12e^{4x}}\right)=$$ $$\lim_{x\to 0}\frac{1-e^{2x}}{-3+3e^{4x}+3x}+5\left(\frac{1}{3+12e^{4\cdot 0}}\right)=$$ $$\lim_{x\to 0}\frac{1-e^{2x}}{-3+3e^{4x}+3x}+5\left(\frac{1}{15}\right)=$$ $$\lim_{x\to 0}\frac{1-e^{2x}}{-3+3e^{4x}+3x}+\frac{5}{15}=$$ $$\lim_{x\to 0}\frac{1-e^{2x}}{-3+3e^{4x}+3x}+\frac{1}{3}=$$ $$\lim_{x\to 0}\frac{\frac{\text{d}}{\text{d}x}\left(1-e^{2x}\right)}{\frac{\text{d}}{\text{d}x}\left(-3+3e^{4x}+3x\right)}+\frac{1}{3}=$$ $$\lim_{x\to 0}-\frac{2e^{2x}}{3(1+4e^{4x})}+\frac{1}{3}=$$ $$-\frac{2e^{2\cdot 0}}{3(1+4e^{4\cdot 0})}+\frac{1}{3}=$$ $$-\frac{2}{15}+\frac{1}{3}=\frac{1}{5}$$