Question

If f(x)=4x^3 ,find lim t➡️0 [ f(2+t) - f(2) ] ÷ t

Answer 1

Answer:

$\bold\red{Value=48}$

Step-by-step explanation:

Given,

$f(x) = 4 {x}^{3}$

Therefore,

$f(2 + t) = 4 {(2 + t)}^{3}$

and,

$f(2) = 4 \times {2}^{3} = 4 \times 8 = 32$

Now, we have to find the

$\lim_{t \to 0} \frac{f(2 + t) - f(2)}{t} \\ \\ = \lim_{t \to 0} \frac{4 {(2 + t)}^{3} - 32 }{t}$

Putting the value of t = 0,

we observe that,

it is limit of 0/0 form

Therefore,

Apply L- Hospital rule

For that,

differentiating the numerator and denominator wrt t,

we get,

$= \lim_{t \to 0} \frac{4 \times 3 {(2 + t)}^{2} - 0}{1} \\ \\ \ = lim_{t \to 0} \: 12 {( 2 + t)}^{2} \\ \\ = 12\lim_{t \to 0} {(2 + t)}^{2}$

Putting the value of t = 0,

we get,

$= 12 \times {2}^{2} \\ \\ = 12 \times 4 \\ \\ = 48$

Hence, the value of given limit is 48