Calculate the limit. lim_(x rarr4)(|x+1||x-4|)/(x^(2)-5x+4)

Factor Denominator: First, let's factor the denominator. x^2 - 5x + 4 = (x - 4)(x - 1)Rewrite Limit: Now, let's rewrite the limit with the factored denominator. lim_(x rarr4)(|x+1||x-4|)/((x - 4)(x - 1))Cancel Absolute Value: Notice that |x-4| and (x-4) will cancel each other out, but we need to be careful because of the absolute value.Split Limit into Cases: Since we're approaching 4 from both sides, |x-4| = x-4 when x > 4 and |x-4| = -(x-4) when x 4 and one for x 4: lim_(x rarr4+)(|x+1|(x-4))/((x - 4)(x - 1)) For x 4: lim_(x rarr4+)(|x+1|)/(x - 1) For x 4: lim_(x rarr4+)(5)/(x - 1) For x 4: lim_(x rarr4+)(5)/(x - 1) = 5/(4 - 1) = 5/3 For x < 4: lim_(x rarr4-)(-5)/(x - 1) = -5/(4 - 1) = -5/3

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Calculate the limit.
lim_(x rarr4)(|x+1||x-4|)/(x^(2)-5x+4)

Calculate the limit. $\newline$ $\lim _{x \rightarrow 4} \frac{|x+1||x-4|}{x^{2}-5 x+4}$

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Q. Calculate the limit.

\newline

\lim _{x \rightarrow 4} \frac{|x+1||x-4|}{x^{2}-5 x+4}

Factor Denominator: First, let's factor the denominator. $\newline$ $x^2 - 5x + 4 = (x - 4)(x - 1)$
Rewrite Limit: Now, let's rewrite the limit with the factored denominator. $\newline$ $\lim_{x \to 4}\frac{|x+1||x-4|}{(x - 4)(x - 1)}$
Cancel Absolute Value: Notice that $|x-4|$ and $(x-4)$ will cancel each other out, but we need to be careful because of the absolute value.
Split Limit into Cases: Since we're approaching $4$ from both sides, $|x-4| = x-4$ when x > 4 and $|x-4| = -(x-4)$ when x < 4.
Simplify Both Cases: So, we can split the limit into two cases, one for x > 4 and one for x < 4. For x > 4: $\lim_{x \to 4^+}\left(\frac{|x+1|(x-4)}{(x - 4)(x - 1)}\right)$ For x < 4: $\lim_{x \to 4^-}\left(\frac{|x+1|(-(x-4))}{(x - 4)(x - 1)}\right)$
Evaluate Absolute Value: Now, let's simplify both cases. $\newline$ For x > 4: $\lim_{x \to 4^+}\left(\frac{|x+1|}{x - 1}\right)$ $\newline$ For x < 4: $\lim_{x \to 4^-}\left(\frac{-|x+1|}{x - 1}\right)$
Find Limits: As $x$ approaches $4$ , $|x+1|$ becomes $|4+1|$ which is $5$ .
Find Limits: As $x$ approaches $4$ , $|x+1|$ becomes $|4+1|$ which is $5$ .So, we have: $\newline$ For x > 4: $\lim_{x \to 4^+}\frac{5}{x - 1}$ $\newline$ For x < 4: $\lim_{x \to 4^-}\frac{-5}{x - 1}$
Find Limits: As $x$ approaches $4$ , $|x+1|$ becomes $|4+1|$ which is $5$ .So, we have: $\newline$ For x > 4: $\lim_{x \to 4^+}\frac{5}{x - 1}$ $\newline$ For x < 4: $\lim_{x \to 4^-}\frac{-5}{x - 1}$ Now, let's find the limits. $\newline$ For x > 4: $4$ $0$ $\newline$ For x < 4: $4$ $2$