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Find lim_(x rarr1)h(x) for h(x)=5x^(3)-6x^(2)+2x-1". "

Find lim⁡x→1h(x) \lim _{x \rightarrow 1} h(x) for\newlineh(x)=5x3−6x2+2x−1.  h(x)=5 x^{3}-6 x^{2}+2 x-1 \text {. }

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Q. Find lim⁡x→1h(x) \lim _{x \rightarrow 1} h(x) for\newlineh(x)=5x3−6x2+2x−1.  h(x)=5 x^{3}-6 x^{2}+2 x-1 \text {. }
  1. Direct Substitution: To find the limit of h(x)h(x) as xx approaches 11, we can directly substitute x=1x = 1 into the function h(x)h(x), since h(x)h(x) is a polynomial and polynomials are continuous everywhere.
  2. Substitute and Calculate: Substitute x=1x = 1 into h(x)h(x):h(1)=5(1)3−6(1)2+2(1)−1h(1) = 5(1)^3 - 6(1)^2 + 2(1) - 1
  3. Simplify Expression: Calculate the value of h(1)h(1):h(1)=5(1)−6(1)+2(1)−1h(1) = 5(1) - 6(1) + 2(1) - 1h(1)=5−6+2−1h(1) = 5 - 6 + 2 - 1
  4. Simplify Expression: Calculate the value of h(1)h(1):h(1)=5(1)−6(1)+2(1)−1h(1) = 5(1) - 6(1) + 2(1) - 1h(1)=5−6+2−1h(1) = 5 - 6 + 2 - 1Simplify the expression:h(1)=5−6+2−1h(1) = 5 - 6 + 2 - 1h(1)=0h(1) = 0

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