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Evaluate the following : limx→2[logx-log2x-2]

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प्रश्न

Evaluate the following :

`lim_(x -> 2) [(logx - log2)/(x - 2)]`

योग
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उत्तर

Put x = 2 + h. Then x – 2 = h and as x → 2, h → 0.

∴ `lim_(x -> 2) [(log x - log 2)/(x - 2)]`

= `lim_("h" -> 0) [(log(2 + "h") - log2)/"h"]`

= `lim_("h" -> 0) (log ((2 + "h")/2))/"h"`

= `lim_("h" -> 0) (log(1 + "h"/2))/(("h"/2) xx 2)`

= `1/2 lim_("h" -> 0) (log(1 + "h"/2))/(("h"/2))`

= `1/2 xx 1   ...[because "h" -> 0, "h"/2 -> 0  "and" lim_(x -> 0) (log(1 + x))/x = 1]`

= `1/2`

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 7: Limits - Miscellaneous Exercise 7.2 [पृष्ठ १५९]

APPEARS IN

बालभारती Mathematics and Statistics (Arts and Science) Part 2 [English] Standard 11 Maharashtra State Board
अध्याय 7 Limits
Miscellaneous Exercise 7.2 | Q II. (12) | पृष्ठ १५९

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