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Question
Evaluate the following :
`lim_(x -> 1) [(2^(2x - 2) - 2^x + 1)/(sin^2 (x - 1))]`
Solution
Put x = 1 + h.
Then x – 1 = h
and as x → 1, h → 0
∴ `lim_(x -> 1) (2^(2x - 2) - 2^x + 1)/(sin^2 (x - 1))`
= `lim_(x -> 1) (2^(2(x - 1)) - 2^x + 1)/(sin^2 (x - 1))`
= `lim_("h" -> 0) (2^(2"h") - 2^(1 + "h") + 1)/(sin^2"h")`
= `lim_("h" -> 0) (2^(2"h") - 2*2^"h" + 1)/(sin^2"h")`
= `lim_("h" -> 0) (2^"h" - 1)^2/(sin^2"h")`
= `lim_("h" -> 0) ((2^"h" - 1)/"h")^2/((sin"h")/"h")^2` ...[∵ h → 0 ∴ h ≠ 0]
= `(lim_("h" -> 0) (2^"h" - 1)/"h")^2/(lim_("h" -> 0) sin"h"/"h")^2`
= `(log 2)^2/(1)^2 ...[∵ lim_(x -> 0) ("a"^x - 1)/x = log "a"]`
= (log 2)2
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