Question

lim x -- 5
e^x-e^5÷x-5​

Answer 1

$\displaystyle \rm \longrightarrow \lim_{ x\to5}{ \dfrac{ {e}^{x} -{e}^{5} }{x - 5} }$

If we put x = 5 in the above expression then we will get 0/0 form. Now , 0/0 is indeterminate form. So, we will apply L'Hospital rule while solving this question. We will differentiate both numerator and denominator seperately.

$\displaystyle \rm \longrightarrow \lim_{ x\to5} \dfrac{ \dfrac{d({e}^{x} -{e}^{5})}{dx} }{ \dfrac{d(x - 5)}{dx} } \\ \\ \\ \displaystyle \rm \longrightarrow \lim_{ x\to5}{ \dfrac{ {e}^{x} -0}{1 - 0} } \\ \\ \\ \displaystyle \rm \longrightarrow \lim_{ x\to5}{ \dfrac{ {e}^{x}}{1} }\\ \\ \\ \displaystyle \rm \longrightarrow {e}^{5}$

Additional Information :-

d(e^x)/dx = e^x

d(x^n)/dx = n x^(n-1)

d(ln x)/dx = 1/x

d(sin x)/dx = cos x

d(cos x)/dx = - sin x

d(tan x)/dx = sec² x

d(sec x)/dx = tan x * sec x

d(cot x)/dx = - cosec²x

d(cosec x)/dx = - cosec x * cot x

Answer 2

$\large\underline\blue{\bold{Given \:  Question :-  }}$

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$\bf \:Evaluate \: \displaystyle \rm \longrightarrow \lim_{ x\to5} \dfrac{ {e}^{x} - {e}^{5} }{x - 5}$

$\large\underline\blue{\bold{Formula \: used:- }}$

$\bf \: \displaystyle \rm \ \lim_{ x\to0} \dfrac{ ( {e}^{x} - 1) }{x } = 1$

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$\huge{AηsωeR} ✍$

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$\bf \:\displaystyle \rm \longrightarrow \lim_{ x\to5} \dfrac{ {e}^{x} - {e}^{5} }{x - 5}$

$\sf \:  Put \: x = 5 + h$

$\sf \:  As \: x⟼5\sf\implies \:h⟼0$

$\bf \:\displaystyle \rm \longrightarrow So, \: \lim_{ x\to5} \dfrac{ {e}^{x} - {e}^{5} }{x - 5} = \bf \:\displaystyle \rm \ \lim_{ h\to0} \dfrac{ {e}^{5 + h} - {e}^{5} }{5 + h - 5}$

$\bf \: = \displaystyle \rm \ \lim_{ h\to0} \dfrac{ {e}^{5} \times {e}^{h} - {e}^{5} }{h }$

$\bf \: = \displaystyle \rm \ \lim_{ h\to0} \dfrac{ {e}^{5} ( {e}^{h} - 1) }{h }$

$\bf \: = \displaystyle \rm \ {e}^{5} \lim_{ h\to0} \dfrac{ ( {e}^{h} - 1) }{h }$

$\sf \:  = {e}^{5} \times 1 = {e}^{5}$

$\large{\boxed{\boxed{\bf{Hence, \bf \:\displaystyle \rm \ \lim_{ x\to5} \dfrac{ {e}^{x} - {e}^{5} }{x - 5} = {e}^{5} }}}}$

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