# What is the equation of the normal line of #f(x)=xln(1/x)-1/x# at #x=5#?

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To find the equation of the normal line at a given point, we need to determine the slope of the tangent line at that point and then find the negative reciprocal of that slope to obtain the slope of the normal line.

First, let's find the derivative of the function f(x) = xln(1/x) - 1/x using the product rule and the chain rule:

f'(x) = (1)(ln(1/x)) + x(1/x)(-1/x^2) - (-1/x^2) = ln(1/x) - 1/x^2 + 1/x^2 = ln(1/x)

Now, let's evaluate the derivative at x = 5 to find the slope of the tangent line:

f'(5) = ln(1/5)

To find the slope of the normal line, we take the negative reciprocal of the slope of the tangent line:

m_normal = -1/f'(5)

Therefore, the equation of the normal line at x = 5 is y - f(5) = m_normal(x - 5).

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The equation of the normal line of ( f(x) = x \ln\left(\frac{1}{x}\right) - \frac{1}{x} ) at ( x = 5 ) is ( y = -\frac{1}{5}x + \frac{3}{5} ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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