Question

➡Hey Guys __!!!


______________


★ write L'Hospital Rule __¿?

★Explane by a example __!!
________________

➡If u Know then give Answer, otherwise Don't give ___!!!!!!!!​

Answer 1

$\Large{\underline{\underline{\mathfrak{\bf{Question}}}}}$

★ write L'Hospital Rule ?

$\Large{\underline{\underline{\mathfrak{\bf{Answer}}}}}$

$\Large{\underline{\underline{\mathfrak{\bf{L'Hospital's\:Rule}}}}}$

L'Hospital's rule is the definitive way to simplify evaluation of limits. It does not directly evaluate limits , but only simplifies evaluation of used appropriately .

It effect , this rule is the ultimate version of "cancellation tricks " applicable in situations where a more down to earth genuine algebraic cancellation may be hidden or invisible .

Suppose we want to calculate ,

$\:\:\:\:\:\:\displaystyle \lim_{x \to a} \dfrac{f(x)}{g(x)}$

Where, the limit a could also be + infinity or - infinity in addition to "ordinary" numbers .

Suppose that either ,

$\displaystyle \lim_{x \to a} f(x)=0\:\:and\:\:\displaystyle \lim_{x \to a} g(x)=0$

Or,

$\displaystyle \lim_{x \to a} f(x)=+infinite\:\:and\:\:\displaystyle \lim_{x \to a} g(x)=-infinite$

Then we can't just "plug in" to evaluate the limit, and these are traditionally called "indeterminate form" .

The unexpected trick that work often is that (amazingly) we are entitled to take the derivative of numerator and denominator .

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

$\Large{\underline{\underline{\mathfrak{\bf{Example}}}}}$

$\sf{\pink{\:\:Find\:\displaystyle \lim_{x \to 0} \dfrac{x}{e^{2x}-1}}}$

$\Large{\underline{\underline{\mathfrak{\bf{Solution}}}}}$

Both numerator and denominator go to 0 , so we are entitle to use L'Hospital's rule:

$\:\:\:\:\:\displaystyle \lim_{x \to 0} \dfrac{x}{e^{2x}-1}\:=\:\displaystyle \lim_{x \to 0} \dfrac{1}{2e^{2x}}$

In the new expression, numerator and denominator are both non zero , when x = 0 ,

so we just plug in 0 to get,

$\mapsto\sf{\pink{\:\displaystyle \lim_{ x \to 0} \dfrac{1}{2e^{2x}}}}$

$\mapsto\sf{\:\dfrac{1}{2e^{2\times0}}}$

$\mapsto\sf{\:\dfrac{1}{2e^{0}}}$

$\mapsto\sf{\:\dfrac{1}{2\times1}}$

$\mapsto\sf{\:\dfrac{1}{2}}$

____________________

$\Large{\underline{\underline{\mathfrak{\bf{Using\:Identity}}}}}$

$\mapsto\sf{\:e^{0}\:=\:1}$

__________________

Answer 2

Answer:

e0 = 1

Step-by-step explanation:

I hope it is helpful to you