If f x x ndash 5 x 5 k for x lt 5 k h for x 5 x 5 x 5 ...........

Mathematics (NDA) limits

If f(x) = (x – 5)/ |x-5| + k, for x<5

= k + h, for x = 5

= x-5 / |x-5| + h, for x >5

And f(x) is continuous at x = 5, then find the value of k and h

1,1

-1,1

1,-1

1,2

We have, lim f(x) = lim (x-5)/ |x-5| + k

x→5(-) x →5(-)

= lim(x-5) / -(x-5) + k

X→5

= -1 + k ……. (1)

Also, limf(x) = lim(x-5) / |x-5| +h

x →5(+) x →5(+)

= lim(x-5) / (x-5) + h

x→5

= 1 + h ……. (2)

Since, given f(x) is continuous at x= 5

Therefore, limf(x) = f (5) = limf(x)

x→5(-) x→5(+)

⇒ -1 + k = k + h = 1 + h

⇒ k = 1, h = -1

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