ISI2015-DCG-58

Question

$\underset{x \to 1}{\lim} \dfrac{x^{16}-1}{\mid x-1 \mid}$ equals

• A. $-1$
• B. $0$
• C. $1$
• D. Does not exist

Comments

d?

Answer 1

Break the limit at x = 1,

i.e calculate LHL and RHL and then see, if they come out to be unequal, then limit doesn’t exist, else limit exist.

Now, for LHL, |x – 1|, if x < 1 then it become -(x – 1)

so, LHL = lim h→0 ((1-h)^16 – 1) / (-(1-h – 1)), on calculating it comes out to be -17(please do a check)

similarily RHL = lim h→0 ((1+h)^16 – 1) / (1+h-1), which comes out to be 15(calculation might be wrong)

clearly, limit doesn’t exist at x = 1.

Option D