QuestionFind if the limit $\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}$ exists for $f(x)=x^{2}+4x$ and calculate its value.

Studdy Solution

STEP 1

Assumptions1. The function $f(x)$ is given as $f(x)=x^{}+4 x$ . We are asked to find the limit of the expression $\frac{f(1+h)-f(1)}{h}$ as $h$ approaches $0$

STEP 2

First, we need to find the value of $f(1+h)$ by substituting $1+h$ into the function $f(x)$ .
$f(1+h) = (1+h)^{2}+4 (1+h)$

STEP 3

Expand the expression for $f(1+h)$ .
$f(1+h) =1 +2h + h^{2} + +h$

STEP 4

implify the expression for $f(1+h)$ .
$f(1+h) = h^{2} +6h +$

STEP 5

Now, we need to find the value of $f(1)$ by substituting $1$ into the function $f(x)$ .
$f(1) =1^{2}+4 (1)$

STEP 6

implify the expression for $f(1)$ .
$f(1) =1 +4 =5$

STEP 7

Substitute the expressions for $f(1+h)$ and $f(1)$ into the limit expression.
$\lim{h \rightarrow0} \frac{f(1+h)-f(1)}{h} = \lim{h \rightarrow0} \frac{(h^{2} +6h +5) -5}{h}$

STEP 8

implify the expression inside the limit.
$\lim{h \rightarrow0} \frac{h^{2} +6h}{h}$

STEP 9

Divide each term in the numerator by $h$ .
$\lim{h \rightarrow} h +6$

STEP 10

As $h$ approaches $0$ , the term $h$ will become $0$ .
$\lim{h \rightarrow0} h +6 =0 +6 =6$ So, the limit exists and its value is $6$ .

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QuestionFind if the limit $\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}$ exists for $f(x)=x^{2}+4x$ and calculate its value.

Studdy Solution

STEP 1

Assumptions1. The function $f(x)$ is given as $f(x)=x^{}+4 x$ . We are asked to find the limit of the expression $\frac{f(1+h)-f(1)}{h}$ as $h$ approaches $0$

STEP 2

First, we need to find the value of $f(1+h)$ by substituting $1+h$ into the function $f(x)$ .
$f(1+h) = (1+h)^{2}+4 (1+h)$

STEP 3

Expand the expression for $f(1+h)$ .
$f(1+h) =1 +2h + h^{2} + +h$

STEP 4

implify the expression for $f(1+h)$ .
$f(1+h) = h^{2} +6h +$

STEP 5

Now, we need to find the value of $f(1)$ by substituting $1$ into the function $f(x)$ .
$f(1) =1^{2}+4 (1)$

STEP 6

implify the expression for $f(1)$ .
$f(1) =1 +4 =5$

STEP 7

Substitute the expressions for $f(1+h)$ and $f(1)$ into the limit expression.
$\lim{h \rightarrow0} \frac{f(1+h)-f(1)}{h} = \lim{h \rightarrow0} \frac{(h^{2} +6h +5) -5}{h}$

STEP 8

implify the expression inside the limit.
$\lim{h \rightarrow0} \frac{h^{2} +6h}{h}$

STEP 9

Divide each term in the numerator by $h$ .
$\lim{h \rightarrow} h +6$

STEP 10

As $h$ approaches $0$ , the term $h$ will become $0$ .
$\lim{h \rightarrow0} h +6 =0 +6 =6$ So, the limit exists and its value is $6$ .

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QuestionFind if the limit lim⁡h→0f(1+h)−f(1)h\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}h→0lim​hf(1+h)−f(1)​ exists for f(x)=x2+4xf(x)=x^{2}+4xf(x)=x2+4x and calculate its value.

Studdy Solution

STEP 1

Assumptions1. The function f(x)f(x)f(x) is given as f(x)=x+4xf(x)=x^{}+4 xf(x)=x+4x . We are asked to find the limit of the expression f(1+h)−f(1)h\frac{f(1+h)-f(1)}{h}hf(1+h)−f(1)​ as hhh approaches 000

STEP 2

First, we need to find the value of f(1+h)f(1+h)f(1+h) by substituting 1+h1+h1+h into the function f(x)f(x)f(x).f(1+h)=(1+h)2+4(1+h)f(1+h) = (1+h)^{2}+4 (1+h)f(1+h)=(1+h)2+4(1+h)

STEP 3

Expand the expression for f(1+h)f(1+h)f(1+h).f(1+h)=1+2h+h2++hf(1+h) =1 +2h + h^{2} + +hf(1+h)=1+2h+h2++h

STEP 4

implify the expression for f(1+h)f(1+h)f(1+h).f(1+h)=h2+6h+f(1+h) = h^{2} +6h +f(1+h)=h2+6h+

STEP 5

Now, we need to find the value of f(1)f(1)f(1) by substituting 111 into the function f(x)f(x)f(x).f(1)=12+4(1)f(1) =1^{2}+4 (1)f(1)=12+4(1)

STEP 6

implify the expression for f(1)f(1)f(1).f(1)=1+4=5f(1) =1 +4 =5f(1)=1+4=5

STEP 7

Substitute the expressions for f(1+h)f(1+h)f(1+h) and f(1)f(1)f(1) into the limit expression.lim⁡h→0f(1+h)−f(1)h=lim⁡h→0(h2+6h+5)−5h\lim{h \rightarrow0} \frac{f(1+h)-f(1)}{h} = \lim{h \rightarrow0} \frac{(h^{2} +6h +5) -5}{h}limh→0hf(1+h)−f(1)​=limh→0h(h2+6h+5)−5​

STEP 8

implify the expression inside the limit.lim⁡h→0h2+6hh\lim{h \rightarrow0} \frac{h^{2} +6h}{h}limh→0hh2+6h​

STEP 9

Divide each term in the numerator by hhh.lim⁡h→h+6\lim{h \rightarrow} h +6limh→h+6

STEP 10

As hhh approaches 000, the term hhh will become 000.lim⁡h→0h+6=0+6=6\lim{h \rightarrow0} h +6 =0 +6 =6limh→0h+6=0+6=6So, the limit exists and its value is 666.

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QuestionFind if the limit $\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}$ exists for $f(x)=x^{2}+4x$ and calculate its value.

Assumptions1. The function $f(x)$ is given as $f(x)=x^{}+4 x$ . We are asked to find the limit of the expression $\frac{f(1+h)-f(1)}{h}$ as $h$ approaches $0$

First, we need to find the value of $f(1+h)$ by substituting $1+h$ into the function $f(x)$ .
$f(1+h) = (1+h)^{2}+4 (1+h)$

Expand the expression for $f(1+h)$ .
$f(1+h) =1 +2h + h^{2} + +h$

implify the expression for $f(1+h)$ .
$f(1+h) = h^{2} +6h +$

Now, we need to find the value of $f(1)$ by substituting $1$ into the function $f(x)$ .
$f(1) =1^{2}+4 (1)$

implify the expression for $f(1)$ .
$f(1) =1 +4 =5$

Substitute the expressions for $f(1+h)$ and $f(1)$ into the limit expression.
$\lim{h \rightarrow0} \frac{f(1+h)-f(1)}{h} = \lim{h \rightarrow0} \frac{(h^{2} +6h +5) -5}{h}$

implify the expression inside the limit.
$\lim{h \rightarrow0} \frac{h^{2} +6h}{h}$

Divide each term in the numerator by $h$ .
$\lim{h \rightarrow} h +6$

As $h$ approaches $0$ , the term $h$ will become $0$ .
$\lim{h \rightarrow0} h +6 =0 +6 =6$ So, the limit exists and its value is $6$ .