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प्रश्न
Find `lim_(x -> 1)` f(x), where `f(x) = {(x^2 -1, x <= 1), (-x^2 -1, x > 1):}`
उत्तर
When x < 1, f(x) = x2 − 1
By keeping the value of x in the function less than 1 and near 1,
| x | 0.9 | 0.99 | 0.999 |
| f(x) | −0.19 | −0.0199 | −0.0019999 |
∴ `lim_("x" → 1^-) f("x") = 0`
When x >1, f(x) = −x2 − 1 ....(i)
If the value of x in the function is kept greater than 1 and close to 1,
| x | 1.1 | 1.01 | 1.0001 |
| f(x) | −2.21 | −2.0201 | −2.002001 |
∴ `lim_("x" → 1^+) f("x") = -2` .....(ii)
From equations (i) and (ii),
`lim_("x" → 1^-) f("x") ≠ lim_("x" → 1^+) f("x")`
∴ Hence, the equation does not exist at x = 1.
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