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प्रश्न
Evaluate the following limits: `lim_("v" -> sqrt(2))[("v"^2 + "v"sqrt(2) - 4)/("v"^2 - 3"v"sqrt(2) + 4)]`
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उत्तर
`lim_("v" -> sqrt(2))[("v"^2 + "v"sqrt(2) - 4)/("v"^2 - 3"v"sqrt(2) + 4)]`
Consider, `"v"^2 + "v"sqrt(2) - 4 = "v"^2 + sqrt(2)"v" - 4`
= `"v"^2 + 2sqrt(2)"v" - sqrt(2)"v" - 4`
= `"v"("v" + 2sqrt(2)) - sqrt(2)("v" + 2sqrt(2))`
= `("v" + 2sqrt(2)) ("v" - sqrt(2))`
`"v"^2 - 3"v" sqrt(2) + 4 = "v"^2 - 3sqrt(2)"v" + 4`
= `"v"^2 - 2sqrt(2)"v" - sqrt(2)"v" + 4`
= `"v"("v" - 2sqrt(2)) - sqrt(2)("v" - 2sqrt(2))`
= `("v" - 2sqrt(2))("v" - sqrt(2))`
∴ `lim_("v" -> sqrt(2))[("v"^2 + "v"sqrt(2) - 4)/("v"^2 - 3"v"sqrt(2) + 4)]`
= `lim_("v" -> sqrt(2)) (("v" + 2sqrt(2))("v" - sqrt(2)))/(("v" - 2sqrt(2))("v" - sqrt(2))`
= `lim_("v" -> sqrt(2)) ("v" + 2sqrt(2))/("v" - 2sqrt(2)) ...[("As" "v" -> sqrt(2)"," "v" ≠ sqrt(2)),(therefore "v" - sqrt(2) ≠ 0)]`
= `(sqrt(2) + 2sqrt(2))/(sqrt(2) - 2sqrt(2)`
= `(3sqrt(2))/(-sqrt(2)`
= – 3
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