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3 years ago

6

Evaluate the following limit.

1 answer:

Rus_ich [418]3 years ago

7 0

Answer:

-191

Step-by-step explanation:

A limit is the value that a function approaches as the input approaches some value.

We say $\displaystyle \lim_{x\rightarrow a}f(x)=L$ if f(x) approaches to L as x approaches to a.

To find: $\displaystyle \lim_{x\rightarrow -4}2x^3-4x^2+2x+9$

Solution:

Let $f(x)=2x^3-4x^2+2x+9$

On putting x = -4 in function $f(x)=2x^3-4x^2+2x+9$ , we get

$\displaystyle \lim_{x\rightarrow -4}2x^3-4x^2+2x+9\\=2(-4)^3-4(-4)^2+2(-4)+9\\=2(-64)-4(16)-8+9\\=-128-64-8+9\\=-128-63\\=-191$

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Marina86 [1]

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See below

Step-by-step explanation:

$\dfrac{d}{dx} (\tan^3 x) = 3\sec^4 x - 3\sec^2 x$

Recall

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$\dfrac{dy}{dx} =n u^{n-1} \sec^2 x \implies \dfrac{dy}{dx} =n \tan^{n-1} \sec^2 x$

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================================================

For the integration,

$\int \sec^4 x\, dx$

considering the previous part, we will use the identity

$\boxed{\sec^2 x - 1= \tan^2x}$

thus

$\int\sec^4x\,dx=\int \sec^2 x(\tan^2x+1)\,dx = \int \sec^2 x \tan^2x+\sec^2 x\,dx$

and

$\int \sec^2 x \tan^2x+\sec^2 x\,dx = \int \sec^2 x \tan^2x\,dx + \int \sec^2 x\,dx$

Considering $u = \tan x$

and then $du=\sec^2x\ dx$

we have

$\int u^2 \, du = \dfrac{u^3}{3}+C$

Therefore,

$\int \sec^2 x \tan^2x\,dx + \int \sec^2 x\,dx = \dfrac{\tan^3 x}{3}+\tan x + C$

$\boxed{\int \sec^4 x\, dx = \dfrac{\tan^3 x}{3}+\tan x + C }$

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