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प्रश्न
f(x) = `{{:((2x^2 - 3x - 2)/(x - 2)",", "if" x ≠ 2),(5",", "if" x = 2):}` at x = 2
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उत्तर
We have, f(x) = `{{:((2x^2 - 3x - 2)/(x - 2)",", "if" x ≠ 2),(5",", "if" x = 2):}` at x = 2.
At x = 2
L.H.L. = `lim_(x -> 2^-) (2x^2 - 3x - 2)/(x - 2)`
= `lim_("h" -> 0) (2(2 - "h")^2 - 3(2 - "h") - 2)/((2 - "h") - 2)`
= `lim_("h" -> 0) (8 + 2"h"^2 - 8"h" - 6 + 3"h" - 2)/(-"h")`
= `lim_("h" -> 0) (2"h"^2 - 5"h")/(-"h")`
= `lim_("h" -> 0) ("h"(2"h" - 5))/(-"h")` = 5
R.H.L. = `lim_(x -> 2^+) (2x^2 - 3x - 2)/(x - 2)`
= `lim_("h" -> 0) (2(2 + "h")^2 - 3(2 + "h") - 2)/((2 + "h") - 2)`
= `lim_("h" -> 0) (8 + 2"h"^2 + 8"h" - 6 - 3"h" - 2)/"h"`
= `lim_("h" -> 0) (2"h"^2 + 5"h")/"h"`
= `lim_("h" -> 0) ("h"(2"h" + 5))/"h"` = 5
Also f(2) = 5 ....(Given)
∴ L.H.L. = R.H.L. = f(2)
So, f(x) is continuous at x = 2.
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