Question

evaluate lim x. 0 e^5x-1/x​

Answer 1

Question :

Evaluate:

$\sf\lim_{x\to0}\dfrac{e^{5x}-1}{x}$

Solution :

$\tt\lim_{x\to0}\dfrac{e^{5x}-1}{x}$

Since 0/0 is of indeterminate form.

Then ,apply L'Hospital's Rule.

$\implies\sf\lim_{x\to0}\dfrac{e^{5x}-1}{x}=\sf\lim_{x\to0}\dfrac{\dfrac{d(e^{5x})}{dx}-\dfrac{d(1)}{dx}}{\dfrac{dx}{dx}}$

$\sf=\lim_{x\to0}\dfrac{\dfrac{e^{5x}}{d(5x)}\times\dfrac{d(5x)}{dx}-0}{\dfrac{dx}{dx}}$

$\sf=\lim_{x\to0}\dfrac{e^{5x}\times5-0}{1}$

$\sf=\lim_{x\to0}\:e^{5x}\times5$

$\sf=e^{0}\times5$

$\sf=1\times5$

$\sf=5$

Therefore ,

$\rm\lim_{x\to0}\dfrac{e^{5x}-1}{x}=5$

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About L'HOSPITAL'S Rule :

If f(a)=g(a)= 0 , then

$\sf\lim_{x\to\:a}\dfrac{f(x)}{g(x)}=\sf\lim_{x\to\:a}\dfrac{f'(x)}{g'(x)}$