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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→06arctan⁡(1+h)−6arctan⁡(1)h \lim _{h \rightarrow 0} \frac{6 \arctan (1+h)-6 \arctan (1)}{h} .\newlineChoose 11 answer:\newline(A) 11\newline(B) 33\newline(C) 66\newline(D) The limit doesn't exist

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Q. Find lim⁡h→06arctan⁡(1+h)−6arctan⁡(1)h \lim _{h \rightarrow 0} \frac{6 \arctan (1+h)-6 \arctan (1)}{h} .\newlineChoose 11 answer:\newline(A) 11\newline(B) 33\newline(C) 66\newline(D) The limit doesn't exist
  1. Identify Limit: Identify the limit to solve. lim⁡h→06arctan⁡(1+h)−6arctan⁡(1)h\lim_{h \rightarrow 0}\frac{6\arctan(1+h)-6\arctan(1)}{h}
  2. Simplify with arctan⁡(1)\arctan(1): Use the fact that arctan⁡(1)=π4\arctan(1) = \frac{\pi}{4} to simplify the expression.lim⁡h→06arctan⁡(1+h)−6(π4)h\lim_{h \rightarrow 0}\frac{6\arctan(1+h)-6\left(\frac{\pi}{4}\right)}{h}
  3. Apply Limit to Terms: Apply the limit to the arctan⁡(1+h)\arctan(1+h) term.lim⁡h→06arctan⁡(1+h)h−lim⁡h→06(π/4)h\lim_{h \to 0}\frac{6\arctan(1+h)}{h} - \lim_{h \to 0}\frac{6(\pi/4)}{h}
  4. Simplify Constant Term: Simplify the second term, as 6(π/4)6(\pi/4) is a constant and its limit as hh approaches 00 is 00. \newlinelim⁡h→06arctan⁡(1+h)h−0\lim_{h \to 0}\frac{6\arctan(1+h)}{h} - 0
  5. Apply L'Hôpital's Rule: Apply L'Hôpital's Rule to the first term since it's in the indeterminate form 0/00/0. \newlinelim⁡h→0(6ddh(arctan⁡(1+h)))\lim_{h \to 0}(6\frac{d}{dh}(\arctan(1+h)))
  6. Differentiate arctan(1+h)\text{arctan}(1+h): Differentiate arctan(1+h)\text{arctan}(1+h) with respect to hh.ddh(arctan(1+h))=11+(1+h)2\frac{d}{dh}(\text{arctan}(1+h)) = \frac{1}{1+(1+h)^2}
  7. Substitute Derivative: Substitute the derivative back into the limit. lim⁡h→0(61+(1+h)2)\lim_{h \to 0}\left(\frac{6}{1+(1+h)^2}\right)
  8. Evaluate Limit: Evaluate the limit by plugging in h=0h = 0.61+(1+0)2=61+12=62=3\frac{6}{1+(1+0)^2} = \frac{6}{1+1^2} = \frac{6}{2} = 3

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