Find lim_(h rarr0)(6arctan(1+h)-6arctan(1))/(h). Choose 1 answer: (A) 1 (B) 3 (C) 6 (D) The limit doesn't exist

Identify Limit: Identify the limit to solve. lim_(h rarr0)(6arctan(1+h)-6arctan(1))/(h)Simplify with arctan(1): Use the fact that arctan(1) = pi/4 to simplify the expression. lim_(h rarr0)(6arctan(1+h)-6(pi/4))/(h)Apply Limit to Terms: Apply the limit to the arctan(1+h) term. lim_(h rarr0)(6arctan(1+h))/(h) - lim_(h rarr0)(6(pi/4))/(h)Simplify Constant Term: Simplify the second term, as 6(pi/4) is a constant and its limit as h approaches 0 is 0. lim_(h rarr0)(6arctan(1+h))/(h) - 0Apply L'Hôpital's Rule: Apply L'Hôpital's Rule to the first term since it's in the indeterminate form 0/0. lim_(h rarr0)(6(d/dh(arctan(1+h))))Differentiate arctan(1+h): Differentiate arctan(1+h) with respect to h. d/dh(arctan(1+h)) = 1/(1+(1+h)^2)Substitute Derivative: Substitute the derivative back into the limit. lim_(h rarr0)(6/(1+(1+h)^2))Evaluate Limit: Evaluate the limit by plugging in h = 0. 6/(1+(1+0)^2) = 6/(1+1^2) = 6/2 = 3

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Find
lim_(h rarr0)(6arctan(1+h)-6arctan(1))/(h).
Choose 1 answer:
(A) 1
(B) 3
(C) 6
(D) The limit doesn't exist

Find $\lim _{h \rightarrow 0} \frac{6 \arctan (1+h)-6 \arctan (1)}{h}$ . $\newline$ Choose $1$ answer: $\newline$ (A) $1$ $\newline$ (B) $3$ $\newline$ (C) $6$ $\newline$ (D) The limit doesn't exist

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Math Problems

Algebra 2

Solve quadratic inequalities

Full solution

Q. Find

\lim _{h \rightarrow 0} \frac{6 \arctan (1+h)-6 \arctan (1)}{h}

\newline

Choose

1

answer:

\newline

(A)

1

\newline

(B)

3

\newline

(C)

6

\newline

(D) The limit doesn't exist

Identify Limit: Identify the limit to solve. $\lim_{h \rightarrow 0}\frac{6\arctan(1+h)-6\arctan(1)}{h}$
Simplify with $\arctan(1)$ : Use the fact that $\arctan(1) = \frac{\pi}{4}$ to simplify the expression. $\lim_{h \rightarrow 0}\frac{6\arctan(1+h)-6\left(\frac{\pi}{4}\right)}{h}$
Apply Limit to Terms: Apply the limit to the $\arctan(1+h)$ term. $\lim_{h \to 0}\frac{6\arctan(1+h)}{h} - \lim_{h \to 0}\frac{6(\pi/4)}{h}$
Simplify Constant Term: Simplify the second term, as $6(\pi/4)$ is a constant and its limit as $h$ approaches $0$ is $0$ . $\newline$ $\lim_{h \to 0}\frac{6\arctan(1+h)}{h} - 0$
Apply L'Hôpital's Rule: Apply L'Hôpital's Rule to the first term since it's in the indeterminate form $0/0$ . $\newline$ $\lim_{h \to 0}(6\frac{d}{dh}(\arctan(1+h)))$
Differentiate $\text{arctan}(1+h)$ : Differentiate $\text{arctan}(1+h)$ with respect to $h$ . $\frac{d}{dh}(\text{arctan}(1+h)) = \frac{1}{1+(1+h)^2}$
Substitute Derivative: Substitute the derivative back into the limit. $\lim_{h \to 0}\left(\frac{6}{1+(1+h)^2}\right)$
Evaluate Limit: Evaluate the limit by plugging in $h = 0$ . $\frac{6}{1+(1+0)^2} = \frac{6}{1+1^2} = \frac{6}{2} = 3$